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Full text of "Beginning and Intermediate Algebra"

Beginning and Intermediate Algebra 



An open source (CC-BY) textbook 
Available for free download at: http://wallace.ccfaculty.org/book/book.htnil 



by Tyler Wallace 



ISBN #978-1-4583-7768-5 

Copyright 2010, Some Rights Reserved CC-BY. 



Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons 
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Special thanks to: My beautiful wife, Nicole Wallace 

who spent countless hours typing problems and 
my two wonderful kids for their patience and 
support during this project 



Another thanks goes to the faculty reviewers who reviewed this text: Donna Brown, Michelle 
Sherwood, Ron Wallace, and Barbara Whitney 



One last thanks to the student reviewers of the text: Eloisa Butler, Norma Cabanas, Irene 
Chavez, Anna Dahlke, Kelly Diguilio, Camden Eckhart, Brad Evers, Lisa Garza, Nickie Hamp- 
shire, Melissa Hanson, Adriana Hernandez, Tiffany Isaacson, Maria Martinez, Brandon Piatt, 
Tim Ries, Lorissa Smith, Nadine Svopa, Cayleen Trautman, and Erin White 



Table of Contents 



Chapter 0: Pre-Algebra 

. 1 Integers 7 

0.2 Fractions 12 

0.3 Order of Operations 18 

0.4 Properties of Algebra 22 

Chapter 1: Solving Linear Equations 

1.1 One-Step Equations 28 

1.2 Two-Step Equations 33 

1.3 General Linear Equations 37 

1.4 Solving with Fractions 43 

1.5 Formulas 47 

1.6 Absolute Value Equations 52 

1.7 Variation 57 

1.8 Application: Number/ Geometry. 64 

1.9 Application: Age 72 

1.10 Application: Distance 79 

Chapter 2: Graphing 

2.1 Points and Lines 89 

2.2 Slope 95 

2.3 Slope-Intercept Form 102 

2.4 Point-Slope Form 107 

2.5 Parallel & Perpendicular Lines. 112 



Chapter 3: Inequalities 

3.1 Solve and Graph Inequalities.. ..118 

3.2 Compound Inequalitites 124 

3.3 Absolute Value Inequalities 128 

Chapter 4: Systems of Equations 

4.1 Graphing 134 

4.2 Substitution 139 

4.3 Addition/Elimination 146 

4.4 Three Variables 151 

4.5 Application: Value Problems 158 

4.6 Application: Mixture Problems. 167 
Chapter 5: Polynomials 

5.1 Exponent Properties 177 

5.2 Negative Exponents 183 

5.3 Scientific Notation 188 

5.4 Introduction to Polynomials 192 

5.5 Multiply Polynomials 196 

5.6 Multiply Special Products 201 

5.7 Divide Polynomials 205 



Chapter 6: Factoring 

6.1 Greatest Common Factor 212 

6.2 Grouping 216 

6.3 Trinomials where a = l 221 

6.4 Trinomials where a^l 226 

6.5 Factoring Special Products 229 

6.6 Factoring Strategy 234 

6.7 Solve by Factoring 237 

Chapter 7: Rational Expressions 

7.1 Reduce Rational Expressions. ...243 

7.2 Multiply and Divide 248 

7.3 Least Common Denominator.... 253 

7.4 Add and Subtract 257 

7.5 Complex Fractions 262 

7.6 Proportions 268 

7.7 Solving Rational Equations 274 

7.8 Application: Dimensional Analysis. ...279 
Chapter 8: Radicals 

8.1 Square Roots 288 

8.2 Higher Roots 292 

8.3 Adding Radicals 295 

8.4 Multiply and Divide Radicals. ..298 

8.5 Rationalize Denominators 303 

8.6 Rational Exponents 310 

8.7 Radicals of Mixed Index 314 

8.8 Complex Numbers 318 



Chapter 9: Quadratics 

9.1 Solving with Radicals 326 

9.2 Solving with Exponents 332 

9.3 Complete the Square 337 

9.4 Quadratic Formula 343 

9.5 Build Quadratics From Roots. ..348 

9.6 Quadratic in Form 352 

9.7 Application: Rectangles 357 

9.8 Application: Teamwork 364 

9.9 Simultaneous Products 370 

9.10 Application: Revenue and Distance. 373 

9.11 Graphs of Quadratics 380 

Chapter 10: Functions 

10.1 Function Notation 386 

10.2 Operations on Functions 393 

10.3 Inverse Functions 401 

10.4 Exponential Functions 406 

10.5 Logarithmic Functions 410 

10.6 Application: Compound Interest. 414 

10.7 Trigonometric Functions 420 

10.8 Inverse Trigonometric Functions. 428 
Answers 438 



Chapter : Pre-Algebra 

0.1 Integers 7 

0.2 Fractions 12 

0.3 Order of Operations 18 

0.4 Properties of Algebra 22 



0.1 



Pre-Algebra - Integers 



Objective: Add, Subtract, Multiply and Divide Positive and Negative 

Numbers. 

The ability to work comfortably with negative numbers is essential to success in 
algebra. For this reason we will do a quick review of adding, subtracting, multi- 
plying and dividing of integers. Integers are all the positive whole numbers, zero, 
and their opposites (negatives). As this is intended to be a review of integers, the 
descriptions and examples will not be as detailed as a normal lesson. 

World View Note: The first set of rules for working with negative numbers was 
written out by the Indian mathematician Brahmagupa. 

When adding integers we have two cases to consider. The first is if the signs 
match, both positive or both negative. If the signs match we will add the num- 
bers together and keep the sign. This is illustrated in the following examples 



Example 1. 



5 + ( — 3) Same sign, add 5 + 3, keep the negative 
— 8 Our Solution 



Example 2. 



— 7 + ( — 5) Same sign, add 7 + 5, keep the negative 
— 12 Our Solution 

If the signs don't match, one positive and one negative number, we will subtract 
the numbers (as if they were all positive) and then use the sign from the larger 
number. This means if the larger number is positive, the answer is positive. If the 
larger number is negative, the answer is negative. This is shown in the following 
examples. 

Example 3. 

— 7 + 2 Different signs, subtract 7 — 2, use sign from bigger number, negative 
— 5 Our Solution 

Example 4. 



4 + 6 Different signs, subtract 6 — 4, use sign from bigger number, positive 
2 Our Solution 



Example 5. 

4 + ( — 3) Different signs, subtract 4 — 3, use sign from bigger number, positive 
1 Our Solution 

Example 6. 

7 + ( — 10) Different signs, subtract 10 — 7, use sign from bigger number, negative 
— 3 Our Solution 

For subtraction of negatives we will change the problem to an addition problem 
which we can then solve using the above methods. The way we change a subtrac- 
tion to an addition is to add the opposite of the number after the subtraction 
sign. Often this method is refered to as "add the opposite." This is illustrated in 
the following examples. 

Example 7. 

8 — 3 Add the opposite of 3 
8 + ( — 3) Different signs, subtract 8 — 3, use sign from bigger number, positive 
5 Our Solution 



Example 8. 



— 4 — 6 Add the opposite of 6 
4 + ( — 6) Same sign, add 4 + 6, keep the negative 
— 10 Our Solution 



Example 9. 



(-4) Add the opposite of -4 
9 + 4 Same sign, add 9 + 4, keep the positive 
13 Our Solution 



Example 10. 



6 — ( — 2) Add the opposite of — 2 

— 6 + 2 Different sign, subtract 6 — 2, use sign from bigger number, negative 
— 4 Our Solution 



Multiplication and division of integers both work in a very similar pattern. The 
short description of the process is we multiply and divide like normal, if the signs 
match (both positive or both negative) the answer is positive. If the signs don't 
match (one positive and one negative) then the answer is negative. This is shown 
in the following examples 



Example 11. 



(4) ( — 6) Signs do not match, answer is negative 
— 24 Our Solution 



Example 12. 



3G 



Signs match, answer is positive 



Example 13. 



4 Our Solution 



2( — 6) Signs match, answer is positive 
12 Our Solution 



Example 14. 



15 



Signs do not match, answer is negative 



5 Our Solution 



A few things to be careful of when working with integers. First be sure not to 
confuse a problem like — 3 — 8 with — 3( — 8). The second problem is a multipli- 
cation problem because there is nothing between the 3 and the parenthesis. If 
there is no operation written in between the parts, then we assume that means we 
are multiplying. The — 3 — 8 problem, is subtraction because the subtraction sep- 
arates the 3 from what comes after it. Another item to watch out for is to be 
careful not to mix up the pattern for adding and subtracting integers with the 
pattern for multiplying and dividing integers. They can look very similar, for 
example if the signs match on addition, the we keep the negative, — 3 + ( — 7) = — 
10, but if the signs match on multiplication, the answer is positive, ( — 3)( — 7) = 
21. 



0.1 Practice - Integers 

Evaluate each expression. 

I) 1-3 2) 4-(-l) 
3)(-6)-(-8) 4)(-6) + 8 
5)(-3)-3 6 )(-8)-(-3) 

7) 3- (-5) 8)7-7 

9)(-7)-(-5) 10)(-4) + (-l) 

II) 3-(-l) 12) (_l) + (-6) 
13)6-3 14) (-8) + (-1) 
15) (-5) + 3 16) (-1)- 8 
17)2-3 18)5-7 

19) (-8) -(-5) 20) (-5) + 7 

21) (-2) + (-5) 22)l + (-l) 

23) 5- (-6) 24) 8- (-1) 

25) (-6) + 3 26) (-3) + (-1) 

27)4-7 28)7-3 

29) (-7) + 7 30) (-3) + (-5) 
Find each product. 

31)(4)(-1) 32)(7)(-5) 

33)(10)(-8) 34)(-7)(-2) 

35)(-4)(-2) 36)(-6)(-l) 

37)(-7)(8) 38)(6)(-l) 

39)(9)(-4) 40)(-9)(-7) 

41)(-5)(2) 42)(-2)(-2) 

43)(-5)(4) 44)(-3)(-9) 



10 



45) (4) (-6) 

Find each quotient. 

4 6 ) J^_ 47) ^ 

48) =^ 49) ^ 



60) 



8 
54 



-6 



20 



50) f 51, 

i 6 <- ,J v io 



52) %- ki\ z^ 



53) 

,- ^ x 80 

54 ) 3g 55) 



-2 



56) t- K7 ^ Z™ 



57) 2 
59)z 10 



rn^ 48 

58) T r X 60 



11 



0.2 

Pre-Algebra - Fractions 

Objective: Reduce, add, subtract, multiply, and divide with fractions. 

Working with fractions is a very important foundation to algebra. Here we will 
briefly review reducing, multiplying, dividing, adding, and subtracting fractions. 
As this is a review, concepts will not be explained in detail as other lessons are. 

World View Note: The earliest known use of fraction comes from the Middle 
Kingdom of Egypt around 2000 BC! 

We always like our final answers when working with fractions to be reduced. 
Reducing fractions is simply done by dividing both the numerator and denomi- 
nator by the same number. This is shown in the following example 

Example 15. 



36 

84 

36 4- 4 9 



84 4- 4 21 



Both numerator and denominator are divisible by 4 
Both numerator and denominator are still divisible by 3 



12 



Our Soultion 



21 4- 3 7 



The previous example could have been done in one step by dividing both numer- 
ator and denominator by 12. We also could have divided by 2 twice and then 
divided by 3 once (in any order). It is not important which method we use as 
long as we continue reducing our fraction until it cannot be reduced any further. 

The easiest operation with fractions is multiplication. We can multiply fractions 
by multiplying straight across, multiplying numerators together and denominators 
together. 



Example 16. 

6 3 

— • — Multiply numerators across and denominators across 



18 
35 



Our Solution 



When multiplying we can reduce our fractions before we multiply. We can either 
reduce vertically with a single fraction, or diagonally with several fractions, as 
long as we use one number from the numerator and one number from the denomi- 
nator. 



Example 17. 

25 32 

— • — Reduce 25 and 55 by dividing by 5. Reduce 32 and 24 by dividing by 8 

24 55 

5 4 

— • — Multiply numerators across and denominators across 

20 

— Our Solution 

33 



Dividing fractions is very similar to multiplying with one extra step. Dividing 
fractions requires us to first take the reciprocal of the second fraction and mul- 
tiply. Once we do this, the multiplication problem solves just as the previous 
problem. 



13 



Example 18. 

21 ^28 

21 6 



16 28 

3 3 

8*4 

_9_ 
32 



Multiply by the reciprocal 

Reduce 21 and 28 by dividing by 7. Reduce 6 and 16 by dividing by 2 

Multiply numerators across and denominators across 

Our Soultion 



To add and subtract fractions we will first have to find the least common denomi- 
nator (LCD). There are several ways to find an LCD. One way is to find the 
smallest multiple of the largest denominator that you can also divide the small 
denomiator by. 



Example 19. 



Find the LCD of 8 and 12 Test multiples of 12 

12?— Can't divide 12 by 8 

8 J 

24 
24? — = 3 Yes! We can divide 24 by I 

8 J 

24 Our Soultion 



Adding and subtracting fractions is identical in process. If both fractions already 
have a common denominator we just add or subtract the numerators and keep the 
denominator. 



Example 20. 



7 3 



10 



Same denominator, add numerators 7 + 3 



Reduce answer, dividing by 2 



Our Solution 



5 1 

While - can be written as the mixed number 1-, in algebra we will almost never 

4 4) to 

use mixed numbers. For this reason we will always use the improper fraction, not 
the mixed number. 



14 



Example 21. 

13 9 

— — — Same denominator, subtract numerators 13 — 9 
6 6 

4 

— Reduce answer, dividing by 2 
6 

2 

— Our Solution 

o 

If the denominators do not match we will first have to identify the LCD and build 
up each fraction by multiplying the numerators and denominators by the same 
number so the denominator is built up to the LCD. 

Example 22. 

5 4 

^ + - LCD is 18. 

6 9 

3-5 4-2 

1 Multiply first fraction by 3 and the second by 2 

3-6 9-2 yy y y 

15 8 

1 Same denominator, add numerators, 15 + 8 

18 18 

— Our Solution 

18 

Example 23. 

\-\ LCDis6 

3 6 

2-2 1 

— — - — — Multiply first fraction by 2, the second already has a denominator of 6 

2-36 

4 1 

— — — Same denominator, subtract numerators, 4 — 1 
6 6 

3 

— Reduce answer, dividing by 3 
6 

— Our Solution 

2 



15 



0.2 Practice - Fractions 



Simplify each. Leave your answer as an improper fraction. 



42 

i~2 

35 

25 

54 
36 

45 

36 

27 

Is 

40 
16 

63 

18 

80 
60 

72 
60 

36 



1) 

3) 
5) 
7) 
9) 

11) 

13) 

15) 
17) 

19 ) 24 

Find each product. 

21) (9)(|) 

23) (2)(-|) 

25)(-2)(f) 

27) (-§)(-£) 
29) (8)(i) 

31) (f)(1) 
33 (2)(|) 
35)(i)(-5 



2)- 


20 


zH 


24 


4 J - 


9~ 


6)- 


JO 

24 


8)- 


J6 

27 


10) 


48 

Is 


12) 


48 
42 


14) 


16 
12 


16) 


72 
48 


18) 


126 
108 


20) 


160 

Tio 



22) ( 


-2)("i) 


24) ( 


-2)(D 


26) ( 


!)& 


28) 


--)(-- 

7^ 8 


30) ( 


-2)(-?) 


32) ( 


--)(-- 
9 ' v 5 


34) ( 


?)(-!) 


36) ( 


l)(f) 



16 



Find each quotient. 



37 
39 
41 
43 
45 
47 
49 
51 



Evaluate each expression. 



53 
55 
57 
59 
61 
63 
65 
67 
69 
71 
73 
75 
77 
79 
81 



13 

T 

2 
" 3 



1 

To 



i+(-i) 



3 


1 


7 


~~ 7 


1] 
"6 


- + l 


3 

5 


+ 1 


2 

5 


+! 



-+(--) 

! + ("!) 
("*) + ! 



(-1) 



15 



8 

7 

15 

~8~ 

15 s 



(-x) + 



(-l)-(-l) 



38 
40 
42 
44 
46 
48 
50 
52 

54 

56 

58 

60 
62 
64 
66 

68 
70 

72 
74 
76 
78 
80 
82 



12 



2-r 



5 . 
3 ' 


7 
5 


10 
~9~~ 





1 . 
6 ~ 


-5 

IT 



13 



5^5 
3 ' 3 



15 



13 



*+<-£> 



15 x 



("2) + ( s 



12 

T 



(-2) + 



i 

2 ~ 
11 



11 

_ l 

" 2 

8 
5 



(-6) + (-|) 



(-i)-(-^) 



(-*)-(-*) 



(-1) 



17 



0.3 

Pre-Algebra - Order of Operations 

Objective: Evaluate expressions using the order of operations, including 
the use of absolute value. 

When simplifying expressions it is important that we simplify them in the correct 
order. Consider the following problem done two different ways: 



Example 24. 



2 + 5 ■ 3 Add First 2 + 5^3 Multiply 

7^3 Multiply 2_+15 Add 

21 Solution 17 Solution 



The previous example illustrates that if the same problem is done two different 
ways we will arrive at two different solutions. However, only one method can be 
correct. It turns out the second method, 17, is the correct method. The order of 
operations ends with the most basic of operations, addition (or subtraction). 
Before addition is completed we must do repeated addition or multiplication (or 
division). Before multiplication is completed we must do repeated multiplication 
or exponents. When we want to do something out of order and make it come first 
we will put it in parenthesis (or grouping symbols). This list then is our order of 
operations we will use to simplify expressions. 

Order of Operations: 

Parenthesis (Grouping) 

Exponents 

Multiply and Divide (Left to Right) 

Add and Subtract (Left to Right) 

Multiply and Divide are on the same level because they are the same operation 
(division is just multiplying by the reciprocal). This means they must be done left 
to right, so some problems we will divide first, others we will multiply first. The 
same is true for adding and subtracting (subtracting is just adding the opposite). 

Often students use the word PEMDAS to remember the order of operations, as 

the first letter of each operation creates the word PEMDAS. However, it is the 

P 

E 
author's suggestion to think about PEMDAS as a vertical word written as: 

AS 
so we don't forget that multiplication and division are done left to right (same 
with addition and subtraction). Another way students remember the order of 
operations is to think of a phrase such as "Please Excuse My Dear Aunt Sally" 
where each word starts with the same letters as the order of operations start with. 

World View Note: The first use of grouping symbols are found in 1646 in the 
Dutch mathematician, Franciscus van Schooten's text, Vieta. He used a bar over 



18 



the expression that is to be evaluated first. So problems like 2(3 + 5) were written 
as 2 • 3+~5. 

Example 25. 

2 + 3(9 zl 4) 2 Parenthesis first 

2 + 3(5) 2 Exponents 

2 + 3(25) Multiply 

2 + 75 Add 

77 Our Solution 

It is very important to remember to multiply and divide from from left to right! 

Example 26. 

3CKJ3 • 2 Divide first (left to right!) 
1(T2 Multiply 
20 Our Solution 

In the previous example, if we had multiplied first, five would have been the 
answer which is incorrect. 

If there are several parenthesis in a problem we will start with the inner most 
parenthesis and work our way out. Inside each parenthesis we simplify using the 
order of operations as well. To make it easier to know which parenthesis goes with 
which parenthesis, different types of parenthesis will be used such as { } and [ j 
and ( ), these parenthesis all mean the same thing, they are parenthesis and must 
be evaluated first. 

Example 27. 

2{8 2 — 7[32 — 4(3_ 2 + 1)]( — 1)} Inner most parenthesis, exponents first 

2{8 2 - 7[32 - 4(9 + 1)] ( - 1)} Add inside those parenthesis 

2{8 2 — 7[32— 4(10)] ( — 1)} Multiply inside inner most parenthesis 

2{8 2 — 7[32 — 40] ( — 1)} Subtract inside those parenthesis 

2 {8 2 -7[-8](-l)} Exponents next 

2{64— 7[ — 8]( — 1)} Multiply left to right, sign with the number 

2{64 + 56( — 1)} Finish multiplying 

2{64 — 56} Subtract inside parenthesis 

2 {8} Multiply 

16 Our Solution 

As the above example illustrates, it can take several steps to complete a problem. 
The key to successfully solve order of operations problems is to take the time to 
show your work and do one step at a time. This will reduce the chance of making 
a mistake along the way. 

19 



There are several types of grouping symbols that can be used besides parenthesis. 
One type is a fraction bar. If we have a fraction, the entire numerator and the 
entire denominator must be evaluated before we reduce the fraction. In these 
cases we can simplify in both the numerator and denominator at the same time. 



Example 28. 

2 4 - ( - 8) ■ 3 
15H-5-1 

16-^8^3 



2 



40 

2 



Exponent in the numerator, divide in denominator 

Multiply in the numerator, subtract in denominator 

Add the opposite to simplify numerator, denominator is done. 
Reduce, divide 



20 Our Solution 

Another type of grouping symbol that also has an operation with it, absolute 
value. When we have absolute value we will evaluate everything inside the abso- 
lute value, just as if it were a normal parenthesis. Then once the inside is com- 
pleted we will take the absolute value, or distance from zero, to make the number 
positive. 



Example 29. 




l + 3|-4 2 -(-8)| + 2|3 + 0-5)^| 


Evaluate absolute values first, exponents 


1 + 3|-16- (-8)| + 2|3 + 25| 


Add inside absolute values 


~T+3J-8|+2|28| 


Evaluate absolute values 


1 + 3(8) +2(28) 


Multiply left to right 


1 + 24 + 2(28) 


Finish multiplying 


l+J24+~56 


Add left to right 


25 + 56 


Add 


81 


Our Solution 



The above example also illustrates an important point about exponents. Expo- 
nents only are considered to be on the number they are attached to. This means 
when we see — 4 2 , only the 4 is squared, giving us — (4 2 ) or — 16. But when the 
negative is in parentheses, such as ( — 5) 2 the negative is part of the number and 
is also squared giving us a positive solution, 25. 



20 



0.3 Practice - Order of Operation 



Solve. 

I) -6-4(-l) 
3) 3+(8) + |4| 

5) 8H-4-2 
7)[-9-(2-5)] + (-6) 

9) -6+(-3-3) 2 ^|3| 

II) 4-2|3 2 -16| 
13)[-l-(-5)]|3 + 2| 



15) 



2 + 4 7 + 2 2 



4-2 + 5-3 

17) [6-2 + 2- (-6)]( -5 



19) 



13-2 



18 



2-(-l) 3 + (-6)-[-l-(-3)] 



21) 6- 

23) 
25) 



-4+(-4)-[-4-(-3)] 
(4 2 + 3 2 )4 5 

2 3 + 4 
18-6+(-4)-[-5(-l)(-5)] 

5 + 3 2 -24-=-6-2 



2) (-6H-6) 3 
4) 5(-5 + 6)-6 2 
6) 7-5 + 6 

8) (-2-2 3 -2)^(-4) 
10)(- 7 -5)-[-2-2-(-6)] 

12)^-5 

14) -3- {3- [-3(2 + 4) -(-2)]} 
16) -4-[2 + 4(-6)-4-|2 2 -5-2| 
18) 2 • ( - 3) + 3 - 6[ - 2 - ( - 1 - 3)] 

5 2 + (-5) 2 



20) 



|4 2 -2 5 |-2-3 

9- 2- (3 -6) 



22) 1 _ ( _ 2 + 1) _ ( _ 3) 

13+(-3) 2 + 4(-3) + l- 



24) 



10- (-6)] 



{[4 + 5] 4- [4 2 - 3 2 (4 - 3) - 8]} + 12 



[5 + 3(2 2 -5)]+ 2 2 -5| 2 



21 



0.4 



Pre-Algebra - Properties of Algebra 



Objective: Simplify algebraic expressions by substituting given values, 
distributing, and combining like terms 

In algebra we will often need to simplify an expression to make it easier to use. 
There are three basic forms of simplifying which we will review here. 

World View Note: The term "Algebra" comes from the Arabic word al-jabr 
which means "reunion". It was first used in Iraq in 830 AD by Mohammad ibn- 
Musa al-Khwarizmi. 

The first form of simplifying expressions is used when we know what number each 
variable in the expression represents. If we know what they represent we can 
replace each variable with the equivalent number and simplify what remains using 
order of operations. 

Example 30. 



p(q + 6) when p = 3 and q = 5 
(3)((5) + 6) 

(3)(H) 
33 



Replace p with 3 and q with 5 
Evaluate parenthesis 
Multiply 
Our Solution 



Whenever a variable is replaced with something, we will put the new number 
inside a set of parenthesis. Notice the 3 and 5 in the previous example are in 
parenthesis. This is to preserve operations that are sometimes lost in a simple 
replacement. Sometimes the parenthesis won't make a difference, but it is a good 
habbit to always use them to prevent problems later. 

Example 31. 

x + zx(S — z) ( — 1 when x = — 6 and z = — 2 Replace all x's with 6 and z's with 2 

(-6) 



(-6) + (-2)(-6)(3-(-2)) 



6+(-2)(-6)(5)(-2) 

-6 + 12(5)(-2) 

-6 + 60(-2) 



Evaluate parenthesis 

Multiply left to right 
Multiply left to right 
Multiply 



- 6 - 120 Subtract 

— 126 Our Solution 



22 



It will be more common in our study of algebra that we do not know the value of 
the variables. In this case, we will have to simplify what we can and leave the 
variables in our final solution. One way we can simplify expressions is to combine 
like terms. Like terms are terms where the variables match exactly (exponents 
included). Examples of like terms would be 3xy and — 7xy or 3a 2 b and 8a 2 b or — 
3 and 5. If we have like terms we are allowed to add (or subtract) the numbers in 
front of the variables, then keep the variables the same. This is shown in the fol- 
lowing examples 



Example 32. 






5x - 


-2y-8x + 7y 


Combine like terms 5x 




— Sx + 5y 


Our Solution 


Example 33. 






8x 2 - 3x + 7 - 


-2x 2 + Ax-S 


Combine like terms 8x 2 




6x 2 + x + 4 


Our Solution 



8xand — 2y + 7y 



2x 2 and — 3x + 4x and 7 — 3 



As we combine like terms we need to interpret subtraction signs as part of the fol- 
lowing term. This means if we see a subtraction sign, we treat the following term 
like a negative term, the sign always stays with the term. 

A final method to simplify is known as distributing. Often as we work with prob- 
lems there will be a set of parenthesis that make solving a problem difficult, if not 
impossible. To get rid of these unwanted parenthesis we have the distributive 
property. Using this property we multiply the number in front of the parenthesis 
by each term inside of the parenthesis. 

Distributive Property: a(b + c) = ab + ac 

Several examples of using the distributive property are given below. 



Example 34. 



Example 35. 



4 (2x — 7) Multiply each term by 4 
8x — 28 Our Solution 



7(5x — 6) Multiply each term by — 7 
-35 + 42 Our Solution 



In the previous example we again use the fact that the sign goes with the number, 
this means we treat the — 6 as a negative number, this gives ( — 7)( — 6) = 42, a 
positive number. The most common error in distributing is a sign error, be very 
careful with your signs! 



23 



It is possible to distribute just a negative through parenthesis. If we have a nega- 
tive in front of parenthesis we can think of it like a — 1 in front and distribute the 
— 1 through. This is shown in the following example. 



Example 36. 



- (4rr -5y + 6) 

l(4x-5y + 6) 

— 4x + 5y — 6 



Negative can be thought of as — 1 
Multiply each term by — 1 
Our Solution 



Distributing through parenthesis and combining like terms can be combined into 
one problem. Order of operations tells us to multiply (distribute) first then add or 
subtract last (combine like terms). Thus we do each problem in two steps, dis- 
tribute then combine. 



Example 37. 



5 + 3(2x-4) 
5 + 6a; -12 

— 7 + 6a; 



Distribute 3, multipling each term 
Combine like terms 5 — 12 
Our Solution 



Example 38. 



3x-2(4x-5) 

3a; -8a; + 10 

-5a: + 10 



Distribute — 2, multilpying each term 
Combine like terms 3x — 8a: 
Our Solution 



In the previous example we distributed — 2, not just 2. This is because we will 
always treat subtraction like a negative sign that goes with the number after it. 
This makes a big difference when we multiply by the — 5 inside the parenthesis, 
we now have a positive answer. Following are more involved examples of dis- 
tributing and combining like terms. 



Distribute 2 into first parenthesis and — 6 into second 
Combine like terms 10a: — 24a; and — 16 — 18 
Our Solution 



Negative (subtract) in middle can be thought of as — 1 
Distribute 4 into first parenthesis, — 1 into second 
Combine like terms 12a: — 2x and — 32 + 7 
Our Solution 



Example 39. 




2(5x-8)- 
10a; -16- 


-6(4a; + 3) 

- 24a; - 18 

- Ux - 34 


Example 40. 





4(3a; - 8) - (2x - 7) 

4(3x-8)-l(2a:-7) 

12a; - 32 - 2a: + 7 

10a; - 25 



24 



0.4 Practice - Properties of Algebra 

Evaluate each using the values given. 

I) p + 1 + q — m; use m = 1, p = 3, g = 4 2) y 2 + y — z; use y = 5,z=l 
3) p-f;usep = 6andg = 5 4) 6±|z» ;use2 , = l^ = 4 
5)c 2 -(a-l);usea = 3andc = 5 6) x + 62 - 4y; usex = 6, ?/ = 4, 2 = 4 

7) 5j + —;useh = 5,j = 4,k = 2 8 ) 5(5 + a ) + 1 + c; usea = 2, 6 = 6, c = 5 

9) — + q;usem = 4,p = 6,q = 6 10) 2 + x — (l 2 ) 3 ;usex = 5, 2 = 4 

II) m + n + m + |-;usem = land n = 2 12) 3 + z — 1 + y — 1; use g/ = 5, 2 = 4 

13) g — p — (g — 1 — 3); use p = 3, g = 6 

14) p + (5 — r) (6 — p) ; use p = 6, g = 5,r = 5 

15) y — [4— y — (z — x)]; usex = 3, y= l,z = 6 

16) 4^ — (x + x — (2 — z)); use x = 3, z = 2 

17) k x 3 2 - (j + k) - 5;use j = 4, k = 5 18) a 3 (c 2 - c); use a = 3, c= 2 

19) zx- (z-^-);usex = 2,z = 6 20) 5 + qp + pq - q; use p = 6, q = 3 

Combine Like Terms 

21) r- 9 + 10 22) -4x + 2-4 

23) n + n 24)46 + 6 + 1 + 76 

25) %v + 7v 26) -x + 8x 

27) -7x-2x 28) -7a -6 + 5 

29) A; -2 + 7 30) -8p + 5p 

31) x - 10 - 6x + 1 32) 1 - lOn - 10 

33) m - 2m 34) 1 - r - 6 

35) 9n - 1 + n + 4 36) - 46 + 96 



25 



Distribute 






37) -8(x-4) 


38 


) 3(8t> + 9) 


39) 8n(n + 9) 


40 


) -(-5 + 9o) 


41) 7k(-k + 6) 


42 


) 10x(l + 2x) 


43) -6(1 + 6a;) 


44 


) -2(n + l) 


45) 8m(5 — m) 


46 


) -2p(9p-l) 


47) -9x(4-x) 


48 


) 4(8n-2) 


49) -96(6-10) 


50 


) -4(l + 7r) 


51) -8n(5 + 10n) 


52 


) 2x(8x-10) 


Simplify. 






53) 9(6 + 10) + 56 


54 


) 4u-7(l-8u) 


55) -3a;(l-4a;)-4x 2 


56 


) -8a; + 9(-9a; + 9) 


57) -4fe 2 -8fc(8fc + 1) 


58 


) -9- 10(1 + 9a) 


59) l-7(5 + 7p) 


60 


) -10(x-2)-3 


61) -10-4(n-5) 


62 


— 6(5 — m) +3m 


63) 4(x + 7)+8(x + 4) 


64 


) -2r(l + 4r)+8r(-r + 4) 


65) -8(n + 6)-8n(n + 8) 


66 


) 9(66 + 5) -46(6 + 3) 


67) 7(7 + Sv) + 10(3 -lOu) 


68 


) -7(4x- 6) +2(10x-10) 


69) 2n( - lOn + 5) - 7(6 - lOn) 


70 


) -3(4 + a)+6a(9a+10) 


71) 5(1 — 6fc) + 10(fc — 8) 


72 


) -7(4x + 3)-10(10x + 10) 


73) (8n 2 -3n)-(5 + 4n 2 ) 


74 


) (7a; 2 -3)-(5x 2 + 6x) 


75) (5p-6) + (l-p) 


76 


) (3x 2 - x) - (7 - 8x) 


77) (2 - Av 2 ) + (3v 2 + 2u) 


78 


) (26-8) + (6-76 2 ) 


79) (4 - 2fc 2 ) + (8 - 2k 2 ) 


80 


) (7a 2 + 7a)- (6a 2 + 4a) 


81) (a: 2 - 8) + (2x 2 - 7) 


82 


) (3-7n 2 ) + (6n 2 + 3) 



26 



Chapter 1 : Solving Linear Equations 

1 . 1 One-Step Equations 28 

1 . 2 Two-Step Equations 33 

1.3 General Linear Equations 37 

1.4 Solving with Fractions 43 

1 . 5 Formulas 47 

1.6 Absolute Value Equations 52 

1 . 7 Variation 57 

1.8 Application: Number and Geometry 64 

1 . 9 Application: Age 72 

1.10 Application: Distance, Rate and Time 79 



27 



1.1 

Solving Linear Equations - One Step Equations 

Objective: Solve one step linear equations by balancing using inverse 
operations 

Solving linear equations is an important and fundamental skill in algebra. In 
algebra, we are often presented with a problem where the answer is known, but 
part of the problem is missing. The missing part of the problem is what we seek 
to find. An example of such a problem is shown below. 

Example 41. 



4X + 16: 



-4 



Notice the above problem has a missing part, or unknown, that is marked by x. If 
we are given that the solution to this equation is — 5, it could be plugged into the 
equation, replacing the x with — 5. This is shown in Example 2. 



Example 42. 










4(-5) + 16 = 


-4 


Multiply4(-5) 




-20 + 16 = 


-4 


Add -20 + 16 




-4 = 


-4 


True! 



Now the equation comes out to a true statement! Notice also that if another 
number, for example, 3, was plugged in, we would not get a true statement as 
seen in Example 3. 



Example 43. 






4(3) + 16 = -4 Multiply 4(3) 




12 + 16 = -4 Add 12 + 16 




28 ^ - 4 False! 



Due to the fact that this is not a true statement, this demonstates that 3 is not 
the solution. However, depending on the complexity of the problem, this "guess 
and check" method is not very efficient. Thus, we take a more algebraic approach 
to solving equations. Here we will focus on what are called "one-step equations" or 
equations that only require one step to solve. While these equations often seem 
very fundamental, it is important to master the pattern for solving these problems 
so we can solve more complex problems. 



28 



Addition Problems 

To solve equations, the general rule is to do the opposite. For example, consider 
the following example. 



Example 44. 



x + 7 = — 5 The 7 is added to the x 
— 7 — 7 Subtract 7 from both sides to get rid of it 
a; = — 12 Our solution! 



Then we get our so 


lution, 


x = — 12. The same proce 


lowing examples. 






Example 45. 






4 + 2 = 8 




7 = 2 + 9 


-4 -4 




-9 -9 


2 = 4 




-2 = 2 
Table 1. Addition Examples 



5 = 8 + 2 
-8-8 



■ x 



Subtraction Problems 

In a subtraction problem, we get rid of negative numbers by adding them to both 
sides of the equation. For example, consider the following example. 



Example 46. 



2 — 5 = 4 The 5 is negative, or subtracted from x 
+ 5+5 Add 5 to both sides 
2 = 9 Our Solution! 



Then we get our solution 2 = 9. The same process is used in each of the following 
examples. Notice that each time we are getting rid of a negative number by 
adding. 



29 



Example 47. 








-Q +x= -2 




-10 = 2-7 


5=-8+2 


+ 6 +6 




+ 7 +7 


+ 8 +8 


x = 4 




— 3 = x 


13 = 2 




Table 2. 


Subtraction Examples 





Multiplication Problems 

With a multiplication problem, we get rid of the number by dividing on both 
sides. For example consider the following example. 



Example 48. 

4x = 20 Variable is multiplied by 4 
4 4 Divide both sides by 4 
x = 5 Our solution! 

Then we get our solution x = 5 



With multiplication problems it is very important that care is taken with signs. If 
x is multiplied by a negative then we will divide by a negative. This is shown in 
example 9. 



Example 49. 

— 5x = 30 Variable is multiplied by — 5 

— 5 — 5 Divide both sides by — 5 

x = — 6 Our Solution! 



The same process is used in each of the following examples. Notice how negative 
and positive numbers are handled as each problem is solved. 



Example 50. 



30 



ix = - 24 - 4x = - 20 



42 -- 


= 7a; 


~T 


~T 


6 = 


-X 



8 8 -4-4 

X — O 37 — o 

Table 3. Multiplication Examples 



Division Problems: 

In division problems, we get rid of the denominator by multiplying on both sides. 
For example consider our next example. 



Example 51. 



Then we get our 


solution 


X 


lowing examples. 






Example 52. 






JL=_ 2 






(_7)JL=-2 


(-7) 




£ = 14 







X 

— = — 3 Variable is divided by 5 

(5)| = -3(5) Multiply both sides by 5 

x = — 15 Our Solution! 



15. The same process is used in each of the fol- 



- = 9 
-4 y 



(8)| = 5(8) (_ 4 )^ = 9(-4) 

x = 40 x = -36 



Table 4. Division Examples 



The process described above is fundamental to solving equations, once this pro- 
cess is mastered, the problems we will see have several more steps. These prob- 
lems may seem more complex, but the process and patterns used will remain the 
same. 

World View Note: The study of algebra originally was called the "Cossic Art" 
from the Latin, the study of "things" (which we now call variables). 



31 



1.1 Practice - One Step Equations 



Solve each equation. 

1) w + 9 = 16 
3) x- 11 = -16 
5) 30 = a + 20 
7) x - 7 = _ 26 
9) 13 = n-5 



11 

13; 

!s; 

17) 
19) 
21) 

23) 

25) 

27; 

29; 

31 

33) 

35) 

37) 

39) 



340 = 

-9: 



17x 



12 
20t; = -160 

340 = 20n 

16x = 320 

— 16 + n = — 

p-8 = -21 

180 = 12x 



13 



20fe : 



200 



14 14 

-7=a+4 

10 = x-4 

13a =-143 

^ = -12 
20 

9+m=-7 



2) 14 = 6 + 3 

4) -14 = a; -18 

6) -l+fc=5 

8) -13 + P= -19 



10 
12 

14 

16 

18 

20 
22 

24 



22 = 16 + m 



4r : 

5 _ 6 
9 — 9 

-20x : 

1 a 



28 



80 



21 = x + 5 



m — 4: 



13 



26) 


3n = 24 


28) 


-17 = — 
12 


30) 


n + 8 = 10 


32) 


v -16 = -30 


34) 


— 15 = x — 16 


36) 


-8A; = 120 


38) 


" 15 = f 


40) 


-19 = ^ 



20 



32 



1.2 

Linear Equations - Two-Step Equations 



Objective: Solve two-step equations by balancing and using inverse 
opperations. 

After mastering the technique for solving equations that are simple one-step equa- 
tions, we are ready to consider two-step equations. As we solve two-step equa- 
tions, the important thing to remember is that everything works backwards! 
When working with one-step equations, we learned that in order to clear a "plus 
five" in the equation, we would subtract five from both sides. We learned that to 
clear "divided by seven" we multiply by seven on both sides. The same pattern 
applies to the order of operations. When solving for our variable x, we use order 
of operations backwards as well. This means we will add or subtract first, then 
multiply or divide second (then exponents, and finally any parentheses or 
grouping symbols, but that's another lesson). So to solve the equation in the first 
example, 

Example 53. 

Ax - 20 = - 8 

We have two numbers on the same side as the x. We need to move the 4 and the 
20 to the other side. We know to move the four we need to divide, and to move 
the twenty we will add twenty to both sides. If order of operations is done back- 
wards, we will add or subtract first. Therefore we will add 20 to both sides first. 
Once we are done with that, we will divide both sides by 4. The steps are shown 
below. 

Start by focusing on the subtract 20 
Add 20 to both sides 
Now we focus on the 4 multiplied by x 
Divide both sides by 4 
x = 3 Our Solution! 



Notice in our next example when we replace the x with 3 we get a true state- 
ment. 

4(3) -20 = - 8 Multiply 4(3) 
12 - 20 = - 8 Subtract 12 - 20 
- 8 = - 8 True! 



33 



Ax- 


-20 = 


= -8 


+ 20+20 


Ax 




= 12 


~4~ 




~4~ 



The same process is used to solve any two-step equations. Add or subtract first, 
then multiply or divide. Consider our next example and notice how the same pro- 
cess is applied. 



Example 54. 

5x + 7 = 7 Start by focusing on the plus 7 

— 7 — 7 Subtract 7 from both sides 

5x =0 Now focus on the multiplication by 5 

5 5 Divide both sides by 5 

x = Our Solution! 



Notice the seven subtracted out completely! Many students get stuck on this 
point, do not forget that we have a number for "nothing left" and that number is 
zero. With this in mind the process is almost identical to our first example. 

A common error students make with two-step equations is with negative signs. 
Remember the sign always stays with the number. Consider the following 
example. 



Example 55. 






4 -2x = 10 




-4 -4 




-2x =6 



Start by focusing on the positive 4 
Subtract 4 from both sides 
Negative (subtraction) stays on the 2x 
^2 ^2 Divideby-2 
x = — 3 Our Solution! 



The same is true even if there is no coefficient in front of the variable. Consider 
the next example. 



Example 56. 



8 — x = 2 Start by focusing on the positive 8 
8 — 8 Subtract 8 from both sides 



- x = — 6 Negative (subtraction) stays on the x 
lx = — 6 Remember, no number in front of variable means 1 



34 



— 1 — 1 Divide both sides by — 1 
x = 6 Our Solution! 



Solving two-step equations is a very important skill to master, as we study 
algebra. The first step is to add or subtract, the second is to multiply or divide. 
This pattern is seen in each of the following examples. 



Example 57. 


— 3a; 


• + 7 = 


= -8 




-7 


-7 




- 3a; = 


= -15 




-3 


-3 




x = 


5 


7- 


- 5x = 


= 17 


-7 




-7 




- 5x = 


10 




-5 


-5 




x = 


-2 



-2 + 9a; 


= 7 


+ 2 


+ 2 


9a; 


= 9 


~9~ 


~9~ 


X 


= 1 


— 5 — Sx = 


= -5 


+ 5 


+ 5 


— 3a; = 


= 


-3 - 


-3 


x = 






8 = 


= 2a; + 10 


-10 


-10 


-2 = 


= 2a; 


~2~ 


~2~ 


-1 = 


X 


-3 = 


!-" 


+ 4 


+ 4 


(5)(1) 


= 1(5) 


5 = 


-X 



Table 5. Two-Step Equation Examples 



As problems in algebra become more complex the process covered here will 
remain the same. In fact, as we solve problems like those in the next example, 
each one of them will have several steps to solve, but the last two steps are a two- 
step equation like we are solving here. This is why it is very important to master 
two-step equations now! 



Example 58. 

3a; 2 + 4 -a; + 6 w + ~ = w V5x-5 + l = a; log 5 (2x-4) = l 

x — 8 x 3 



World View Note: Persian mathematician Omar Khayyam would solve alge- 
braic problems geometrically by intersecting graphs rather than solving them 
algebraically. 



3.") 



1.2 Practice - Two-Step Problems 



Solve each equation. 



1)5 + ^ = 4 



3) 102 = -7r + 4 
5) -8n + 3 = -77 
7) = - 6v 



9) 
11 
13 
15 
17 

19 
21 
23 
25 
27 
29 
31 
33 
35 
37 
39 



6 







7- 



-12 + 3x = 
24 = 2n - 8 
2 = -12 + 2r 

1 + 7 = 10 
152 = 8n + 64 
-16 = 8a + 64 
56 + 8k = 64 
-2x + 4 = 22 

- 20 = 4p + 4 
-5=3+? 
J-6=-5 

- 40 = 4n - 32 
87 = 3 - 7v 
-x + l = -ll 



2) 


-2 = -2m + 12 


4): 


27 = 21-3:c 


6) 


-4-6=8 


8) 


-2+f=4 


10) 


-5=f-l 


12) 


-6 = 15 + 3p 


14) 


-5m + 2 = 27 


16) 


-37 = 8 + 3x 


18) 


-8 + ^ = -7 


20) 


Y-8=-8 


22) 


-H = -8 + | 


24) 


-2a: -3 = -29 


26) 


-4-3n = -16 


28) 


67 = 5m — 8 


30) 


9 = 8 + f 

6 


32) 


-_1 = _ 2 

4 


34) 


_ 80 = Ax - 28 


36) 


33 = 3o + 3 


38) 


3x — 3 = — 3 


40) 


4 + | = l 



36 



1.3 

Solving Linear Equations - General Equations 

Objective: Solve general linear equations with variables on both sides. 

Often as we are solving linear equations we will need to do some work to set them 
up into a form we are familiar with solving. This section will focus on manipu- 
lating an equation we are asked to solve in such a way that we can use our pat- 
tern for solving two-step equations to ultimately arrive at the solution. 

One such issue that needs to be addressed is parenthesis. Often the parenthesis 
can get in the way of solving an otherwise easy problem. As you might expect we 
can get rid of the unwanted parenthesis by using the distributive property. This is 
shown in the following example. Notice the first step is distributing, then it is 
solved like any other two-step equation. 

Example 59. 

4(2x — 6) = 16 Distribute 4 through parenthesis 

8x — 24 = 16 Focus on the subtraction first 

+ 24 + 24 Add 24 to both sides 

8a; = 40 Now focus on the multiply by 8 

8 8 Divide both sides by 8 

x = 5 Our Solution! 



Often after we distribute there will be some like terms on one side of the equa- 
tion. Example 2 shows distributing to clear the parenthesis and then combining 
like terms next. Notice we only combine like terms on the same side of the equa- 
tion. Once we have done this, our next example solves just like any other two-step 
equation. 



Example 60. 



3(2a; — 4) + 9 = 15 Distribute the 3 through the parenthesis 

6x — 12 + 9 = 15 Combine like terms, — 12 + 9 

6x — 3 = 15 Focus on the subtraction first 

+ 3 +3 Add 3 to both sides 

6x = 18 Now focus on multiply by 6 



37 



6 6 Divide both sides by 6 
x = 3 Our Solution 

A second type of problem that becomes a two-step equation after a bit of work is 
one where we see the variable on both sides. This is shown in the following 
example. 

Example 61. 

4x-6 = 2x + 10 

Notice here the x is on both the left and right sides of the equation. This can 
make it difficult to decide which side to work with. We fix this by moving one of 
the terms with x to the other side, much like we moved a constant term. It 
doesn't matter which term gets moved, 4x or 2x, however, it would be the 
author's suggestion to move the smaller term (to avoid negative coefficients). For 
this reason we begin this problem by clearing the positive 2x by subtracting 2x 
from both sides. 



10 Notice the variable on both sides 
Subtract 2x from both sides 
Focus on the subtraction first 
Add 6 to both sides 
Focus on the multiplication by 2 
Divide both sides by 2 
Our Solution! 



The previous example shows the check on this solution. Here the solution is 
plugged into the x on both the left and right sides before simplifying. 

Example 62. 

4(8) - 6 = 2(8) + 10 Multiply 4(8) and 2(8) first 
32-6 = 16 + 10 Add and Subtract 
26 = 26 True! 

The next example illustrates the same process with negative coefficients. Notice 
first the smaller term with the variable is moved to the other side, this time by 
adding because the coefficient is negative. 

38 



4x- 


-6 = 


2x- 


-2x 


— 


2x 


2x 


-6 = 


= 10 




+ 6+6 




2x- 


= 16 




~2 


~ ~2~ 




X 


= 8 



Example 63. 






- 3x + 9 = 


= 6a; - 


-27 


+ 3x +3x 




9 = 


-9x- 


-27 


+ 27 


+ 27 


36 = 


= 9x 




~9~ 


~9~ 




4: 


= x 





Notice the variable on both sides, — 3x is smaller 

Add 3x to both sides 

Focus on the subtraction by 27 

Add 27 to both sides 

Focus on the mutiplication by 9 

Divide both sides by 9 

Our Solution 



Linear equations can become particularly intersting when the two processes are 
combined. In the following problems we have parenthesis and the variable on both 
sides. Notice in each of the following examples we distribute, then combine like 
terms, then move the variable to one side of the equation. 



Example 


64. 








2(x- 


-5) + Sx = 


= a; + 18 




2x- 


-10 + 3a; = 


= a; + 18 






5a; -10 = 


= a; + 18 






- x — 


X 






Ax -10 = 


= 18 






+ 10 + 10 






4a; = 


= 28 






~2~ 


~4~ 






X 


= 7 



Distribute the 2 through parenthesis 
Combine like terms 2a; + 3x 
Notice the variable is on both sides 
Subtract x from both sides 
Focus on the subtraction of 10 
Add 10 to both sides 
Focus on multiplication by 4 
Divide both sides by 4 
Our Solution 



Sometimes we may have to distribute more than once to clear several parenthesis. 
Remember to combine like terms after you distribute! 



Example 65. 



3(4a; — 5) — 4(2a; + 1) = 5 Distribute 3 and — 4 through parenthesis 

12a; — 15 — 8a; — 4 = 5 Combine like terms 12a; — 8a; and — 15 — 4 

4a; — 19 =5 Focus on subtraction of 19 

+ 19 + 19 Add 19 to both sides 

4x = 24 Focus on multiplication by 4 



39 



4 4 Divide both sides by 4 
x = 6 Our Solution 



This leads to a 5-step process to solve any linear equation. While all five steps 
aren't always needed, this can serve as a guide to solving equations. 

1. Distribute through any parentheses. 

2. Combine like terms on each side of the equation. 

3. Get the variables on one side by adding or subtracting 

4. Solve the remaining 2-step equation (add or subtract then multiply or 
divide) 

5. Check your answer by plugging it back in for x to find a true statement. 

The order of these steps is very important. 

World View Note: The Chinese developed a method for solving equations that 
involved finding each digit one at a time about 2000 years ago! 

We can see each of the above five steps worked through our next example. 
Example 66. 



4(2a; — 6) + 9 = 3(x — 7) + 8x Distribute 4 and 3 through parenthesis 

Combine like terms — 24 + 9 and 3x + 8x 
Notice the variable is on both sides 
Subtract 8a; from both sides 
Focus on subtraction of 21 
Add 21 to both sides 
Focus on multiplication by 3 
Divide both sides by 3 
Our Solution 



•X 


- 24 + 9 = 


-3x — 


21 + 8a; 




8a; -15 = 


= lla; 


-21 




8x 


-8x 






-15 = 


= Sx - 


-21 




+ 21 


+ 21 




6 = 


= 3a; 






~3~ 


~3~ 






2 = 


= x 





Check: 



4 [2 (2) - 6] + 9 = 3 [(2) - 7] + 8(2) Plug 2 in for each a;. Multiply inside parenthesis 
4 [4 — 6] +9 = 3[ — 5] +8(2) Finish parentesis on left, multiply on right 



40 



4[ — 2] +9 = — 15 + 8(2) Finish multiplication on both sides 

_8 + 9 = - 15 + 16 Add 

1 = 1 True! 



When we check our solution of x = 2 we found a true statement, 1 = 1. Therefore, 
we know our solution x = 2 is the correct solution for the problem. 

There are two special cases that can come up as we are solving these linear equa- 
tions. The first is illustrated in the next two examples. Notice we start by dis- 
tributing and moving the variables all to the same side. 



Example 67. 






3(2x- 


- 5) = 6x - 


-15 


6a: - 


- 15 = 6a; - 


-15 


— Qx 


— Qx 





Distribute 3 through parenthesis 
Notice the variable on both sides 
Subtract 6x from both sides 
— 15 = — 15 Variable is gone! True! 

Here the variable subtracted out completely! We are left with a true statement, 
— 15 = — 15. If the variables subtract out completely and we are left with a true 
statement, this indicates that the equation is always true, no matter what x is. 
Thus, for our solution we say all real numbers or R. 



Example 


68. 










2(3x- 


-5)- 


-Ax = 


= 2x + 7 




6x - 


-10- 


-4x = 


= 2x + 7 






2x- 


-10 = 


= 2x + 7 






-2x 




2x 



Distribute 2 through parenthesis 
Combine like terms 6x — 4x 
Notice the variable is on both sides 
Subtract 2x from both sides 
— 10 =£ 7 Variable is gone! False! 

Again, the variable subtracted out completely! However, this time we are left with 
a false statement, this indicates that the equation is never true, no matter what x 
is. Thus, for our solution we say no solution or 0. 



41 



1.3 Practice - General Linear Equations 



Solve each equation. 

I) 2- (-3a- 8) = 1 

3) _5(_4 + 2 v ) = -50 

5) 66 = 6(6 + 5x) 

7) o = -8(p-5) 

9) -2 + 2(8x-7) = -16 

II) -21x + 12 = -6-3x 
13) -l-7m = -8m + 7 
15) l-12r = 29-8r 

17) 20 -76 = -126 + 30 

19) -32-24v = 34-2v 

21) -2-5(2-4m)=33 + 5m 

23) -4n+ll = 2(l-8n)+3n 

25) -6v-29 = -4v-5(v + l) 

27) 2(4x-4) = -20-4x 

29) -a-5(8a-l) = 39-7a 

31) _57 = -(-p + l) + 2(6 + 8p) 

33) -2(m-2) + 7(m-8) = -67 

35) 50 = 8(7 + 7r)-(4r + 6) 

37) -8(n-7) + 3(3n-3)=41 

39) -61 = -5(5r-4)+4(3r-4) 

41) -2(8n-4) = 8(l-n) 

43) -3{-7v + 3) + 8v = 5v-4:(l-6v) 

45) -7(x-2) = -4-6(a;-l) 

47) -6(8fc + 4) = -8(6£; + 3)-2 

49) -2(l-7p) = 8(p-7) 



2) 2(-3n + 8) = -20 
4) 2-8(-4 + 3x)=34 
6) 32 = 2-5(-4n + 6) 

8) _55 = 8 + 7(£;-5) 



io; 


-(3-5n) = 12 


12; 


-3n-27=-27-3n 


14; 


56p-48 = 6p + 2 


16; 


4 + 3x = -12x + 4 


is; 


-16n + 12 = 39-7n 


20; 


17-2^ = 35-82; 


22; 


- 25 - 7a; = 6(2x - 1) 


24; 


-7(1 + 6) = -5-56 


26; 


-8(8r-2) = 3r + 16 


28; 


- 8n - 19 = - 2(8n - 3) + 3n 


30; 


_ 4 + 4k = 4(8k - 8) 


32; 


16 = -5(l-6x)+3(6x + 7) 


34; 


7 = 4(n-7) + 5(7n + 7) 


36; 


- 8(6 + 6x) + 4( - 3 + 6x) = - 12 


38; 


-76 = 5(1 + 36) +3(36 -3) 


40; 


-6(x-8)-4(x-2) = -4 


42; 


- 4(1 + a) = 2a -8(5 + 3a) 


44; 


-6(x -3) + 5 = -2- 5(x -5) 


46; 


- (n + 8) + n = - 8n + 2(4n - 4) 


48; 


-5(x + 7)=4(-8x-2) 


50; 


8(-8n + 4)=4(-7n + 8) 



42 



1.4 

Solving Linear Equations - Fractions 



Objective: Solve linear equations with rational coefficients by multi- 
plying by the least common denominator to clear the fractions. 

Often when solving linear equations we will need to work with an equation with 
fraction coefficients. We can solve these problems as we have in the past. This is 
demonstrated in our next example. 



Example 69. 



—x — — = — Focus on subtraction 
4 2 6 



7 7 7 

+ — +— Add — to both sides 
2 2 2 



5 7 

Notice we will need to get a common denominator to add - + -. Notice we have a 

7 / 3 \ 21 

common denominator of 6. So we build up the denominator, -I - I = — , and we 
can now add the fractions: 

3 21 5 

—x — — = — Same problem, with common denominator 6 

4 6 6 

21 21 21 

-\ 1 Add -— to both sides 

6 6 6 

3 26 _, . 26, 13 
—x = -— Reduce — r to — - 

4 6 6 3 

3 13 3 

— x = — Focus on multiplication by — 

We can get rid of - by dividing both sides by -. Dividing by a fraction is the 
same as multiplying by the reciprocal, so we will multiply both sides by -. 

( - \-x = —( — j Multiply by reciprocal 

52 
x = — Our solution! 

9 

While this process does help us arrive at the correct solution, the fractions can 
make the process quite difficult. This is why we have an alternate method for 
dealing with fractions - clearing fractions. Clearing fractions is nice as it gets rid 
of the fractions for the majority of the problem. We can easily clear the fractions 

43 



by finding the LCD and multiplying each term by the LCD. This is shown in the 
next example, the same problem as our first example, but this time we will solve 
by clearing fractions. 



Example 70. 



4 2 6 

(12)3 (12)7 _ (12)5 

4 2 6 



3 7 5 
x — — = — LCD = 12, multiply each term by 12 



Reduce each 12 with denominators 



(3)3x-(6)7 = (2)5 Multiply out each term 

9a; — 42 = 10 Focus on subtraction by 42 
+ 42 + 42 Add 42 to both sides 

9x = 52 Focus on multiplication by 9 
9 9 Divide both sides by 9 
Our Solution 



52 
9 



The next example illustrates this as well. Notice the 2 isn't a fraction in the ori- 
gional equation, but to solve it we put the 2 over 1 to make it a fraction. 



Example 71. 



2 3 1 

— x — 2 = — x + — LCD = 6, multiply each term by 6 



(6)2 ,. (6)2 _ (6)3^ , (6)1 



(2)2; 



1 2 X ' 6 


ixeuu.ce u wim eacii ueiiumin; 


-(6)2 = (3)3x + (l)l 


Multiply out each term 


4x-12 = 9x + l 


Notice variable on both sides 


— Ax —Ax 


Subtract Ax from both sides 


-12 = 5x + l 


Focus on addition of 1 


-1 - 1 


Subtract 1 from both sides 


— 13 = bx 


Focus on multiplication of 5 


~5~ ~5~ 


Divide both sides by 5 


13 


Our Solution 



We can use this same process if there are parenthesis in the problem. We will first 
distribute the coefficient in front of the parenthesis, then clear the fractions. This 
is seen in the following example. 



44 



Example 72. 

Q / JT A \ Q 

—I — x + — 1 = 3 Distribute — through parenthesis, reducing if possible 
2 y 9 27 J 2 

5 2 

— x + — = 3 LCD = 18, multiply each term by 18 

- — — x + - — — = - — — Reduce 18 with each denominator 
6 9 9 

(3)5x+ (2)2= (18)3 Multiply out each term 

15x + 4 = 54 Focus on addition of 4 

— 4—4 Subtract 4 from both sides 



15x = 50 Focus on multiplication by 15 



15 15 Divide both sides by 15. Reduce on right side. 
10 
3 



x = —- Our Solution 



While the problem can take many different forms, the pattern to clear the frac- 
tion is the same, after distributing through any parentheses we multiply each term 
by the LCD and reduce. This will give us a problem with no fractions that is 
much easier to solve. The following example again illustrates this process. 



Distribute — , reduce if possible 
LCD = 4, multiply each term by 4. 

Reduce 4 with each denominator 

Multiply out each term 
Combine like terms 8 — 14 
Notice variable on both sides 
Subtract x from both sides 
Focus on subtraction by 2 
Add 2 to both sides 
Focus on multiplication by 2 
Divide both sides by 2 
Our Solution 

World View Note: The Egyptians were among the first to study fractions and 
linear equations. The most famous mathematical document from Ancient Egypt 
is the Rhind Papyrus where the unknown variable was called "heap" 



Example 73 










3 1 


= 3 ( 4 X + 6) "2 






3 


1 1 o 7 

-2 = 4* + 2 -2 


(4)3 x 


(4)1 
2 


(4)1, 


, , (4)2 (4)7 


4 X 


4 ■*■' 1 2 


(1)32 


-(2)1 = 


= (1)1:2 + (4)2 -(2)7 






3x 


-2 = 2 + 8-14 

3x — 2 = x — 6 

— x — X 

2x-2 = -6 

+ 2 +2 

2x = -4 

~2~ ~2~ 

x=-2 



45 



1.4 Practice - Fractions 



Solve each equation. 

3) = 



21 
20 



5/ 6x 



3 5 



5) -r — -111 - 



113 
~24~ 



7) 



635 _ _ 5 / _ 
~T2~ ~ 2^~ 

9 11 



X) 



9) 26 



n)|(^+i)=| 



13) - a -J(-Jo + i; 



19 



55 
16 



15) 

17 ) 9 

19) - 

21) - 



5/3 5x 

K0P-3, 



2 V 2 J 



4 / 4 



5 5/ 3n 



8 4 
11 , 3 



6 = 1(6- 



23) -(■ 



2 X 2> 



3 
' 2 



OK x 45 . 3 7 

25) - + -n = -n 



19 
16 



27)>+|; 



7 19 

4 U "T 



29)f + |x = |(|x + l) 



.7; 



2) 


1 3 t._L 3 
2 - 2 K+ 2 


4) 


3 8 29 
2 n 3 ~~ 12 


6) 


11 . 3 163 

— — r = — 

4 ' 4 32 


8) 


16 4/5 . x 
-^ = -3(3+") 


10) 


3 7 9 
2"4 V= -8 


12) 


41 5/ 2x 1 
9 ~~ 2^ X+ 3> 3 X 


14) 


l<-!* + l)-S*— i 


16) 


1/2 3n 7 83 

~~ 2 V3 X ~~ 4V ~~ 2^ ~~ ~~ 24 


18) 


2/ . 9x 10 53 
3( m + l)-T = -l8 


20) 


1 4 5/ 7x 
12 _ 3 X+ 3^ X 4> 


22) 


7 4 3 . / . 3 
6-3 n = -2 n + 2(^+2 


24) 


149 11 _ 7 5/ 
16 3~ r_ ~~ 4 r ~~ 4 v 


26) 


7/5 . lx 11 .25 
-2-(3 a +3) = T°+T 


28) 


8 1 4 2/ 
~~ 3 ~~ 2 X ~ ~ 3 X ~ 3 v 


30) 


_ n + _ = 2(-n + -) 



13 



r + r 



13 



-x- 



46 



1.5 

Solving Linear Equations - Formulas 

Objective: Solve linear formulas for a given variable. 

Solving formulas is much like solving general linear equations. The only difference 
is we will have several varaibles in the problem and we will be attempting to solve 
for one specific variable. For example, we may have a formula such as A = nr 2 + 
7rrs (formula for surface area of a right circular cone) and we may be interested in 
solving for the varaible s. This means we want to isolate the s so the equation has 
s on one side, and everything else on the other. So a solution might look like s = 

. This second equation gives the same information as the first, they are 

algebraically equivalent, however, one is solved for the area, while the other is 
solved for s (slant height of the cone). In this section we will discuss how we can 
move from the first equation to the second. 

When solving formulas for a variable we need to focus on the one varaible we are 
trying to solve for, all the others are treated just like numbers. This is shown in 
the following example. Two parallel problems are shown, the first is a normal one- 
step equation, the second is a formula that we are solving for x 

Example 74. 

3x = 12 wx = z In both problems, :r is multiplied by something 

3 3 ~W ~W~ To isolate the x we divide by 3 or w . 

x = 4 x = — Our Solution 

w 

We use the same process to solve 3x = 12 for x as we use to solve wx = z for x. 
Because we are solving for x we treat all the other variables the same way we 
would treat numbers. Thus, to get rid of the multiplication we divided by w. This 
same idea is seen in the following example. 

Example 75. 

m + n = p forn Solving for n, treat all other variables like numbers 
— m — m Subtract m from both sides 



n = p — m Our Solution 

As p and m are not like terms, they cannot be combined. For this reason we leave 
the expression as p — m. This same one-step process can be used with grouping 
symbols. 

47 



Example 76. 

a(x — y) = b for a Solving for a , treat (x — y) like a number 
(x — y) (x — y) Divide both sides by (x — y) 

a = Our Solution 

x-y 

Because (x — y) is in parenthesis, if we are not searching for what is inside the 
parenthesis, we can keep them together as a group and divide by that group. 
However, if we are searching for what is inside the parenthesis, we will have to 
break up the parenthesis by distributing. The following example is the same for- 
mula, but this time we will solve for x. 

Example 77. 

a(x — y) = b forx Solving for x, we need to distribute to clear parenthesis 

ax — ay = b This is a two — step equation , ay is subtracted from our x term 

-\- ay -\- ay Add ay to both sides 

ax = b + ay Thesismultipiedby a 

~a~ a Divide both sides by a 

x = Our Solution 

a 

Be very careful as we isolate x that we do not try and cancel the a on top and 
bottom of the fraction. This is not allowed if there is any adding or subtracting in 
the fraction. There is no reducing possible in this problem, so our final reduced 
answer remains x = — — . The next example is another two-step problem 

Example 78. 

y = mx + b for m Solving for m , focus on addition first 

— b — b Subtract b from both sides 

y — b = mx mismultipiedbyrc. 

x ~x~ Divide both sides by x 

= m Our Solution 

x 

It is important to note that we know we are done with the problem when the 
variable we are solving for is isolated or alone on one side of the equation and it 
does not appear anywhere on the other side of the equation. 

The next example is also a two-step equation, it is the problem we started with at 
the beginning of the lesson. 

48 



Example 79. 

A = 7rr 2 + 7rrs for s Solving for s, focus on what is added to the term with 5 
— izr 2 — izr 2 Subtract irr 2 from both sides 

A — irr 2 = 7rrs sis multipied by nr 

izr ~7fr~ Divide both sides by irr 

= s Our Solution 

Again, we cannot reduce the irr in the numerator and denominator because of the 
subtraction in the problem. 

Formulas often have fractions in them and can be solved in much the same way 
we solved with fractions before. First identify the LCD and then multiply each 
term by the LCD. After we reduce there will be no more fractions in the problem 
so we can solve like any general equation from there. 



Example 80. 



2/71 

h = form To clear the fraction we use LCD = n 



n 



(n)h = - — Multiply each term by n 

nh = 2m Reduce n with denominators 

2 2 Divide both sides by 2 

-— = m Our Solution 

The same pattern can be seen when we have several fractions in our problem. 

Example 81. 

a c 

— + — = e for a To clear the fraction we use LCD = b 

b b 

-^ — h -^— = e (b) Multiply each term by b 

a + c = eb Reduce b with denominators 

— c — c Subtract c from both sides 

a = eb — c Our Solution 

Depending on the context of the problem we may find a formula that uses the 
same letter, one capital, one lowercase. These represent different values and we 
must be careful not to combine a capital variable with a lower case variable. 

Example 82. 

A 



, for 6 Use LCD (2 — b) as o group 



49 





2- 


b 


(2- 


- b)a 


= A 


2a- 


- ab 


= A 


-2a 




2a 


— ab = 


= A- 


-2a 


— a 


— 


a 


h- 


A- 


-2a 



(2-b)a = ^- — *f— Multiply each term by (2 -6) 



reduce (2 — b) with denominator 
Distribute through parenthesis 
Subtract 2a from both sides 
The b is multipied by — a 
Divide both sides by — a 

Our Solution 



Notice the A and a were not combined as like terms. This is because a formula 
will often use a capital letter and lower case letter to represent different variables. 
Often with formulas there is more than one way to solve for a variable. The next 
example solves the same problem in a slightly different manner. After clearing the 
denominator, we divide by a to move it to the other side, rather than distributing. 

Example 83. 

A 

for b Use LCD = (2 — 6) as a group 





a>= t. — r 
2-b 


(2- 


m ( 2 - b ) A 
~ b)a= 2-b 




(2-b)a = A 



Multiply each term by (2 — 6) 

Reduce (2 — 6) with denominator 
a ~a~ Divide both sides by a 

2 — b = — Focus on the positive 2 

a 

2 —2 Subtract 2 from both sides 



^4 

— b = 2 Still need to clear the negative 

a 

A 

( — !)(-&) = ( — 1) 2( - 1) Multiply (or divide) each term by 



a 

A 

a 



Our Solution 



Both answers to the last two examples are correct, they are just written in a dif- 
ferent form because we solved them in different ways. This is very common with 
formulas, there may be more than one way to solve for a varaible, yet both are 
equivalent and correct. 

World View Note: The father of algebra, Persian mathematician Muhammad 
ibn Musa Khwarizmi, introduced the fundamental idea of blancing by subtracting 
the same term to the other side of the equation. He called this process al-jabr 
which later became the world algebra. 



.-)() 



1.5 Practice - Formulas 



Solve each of the following equations for the indicated variable. 

1) ab = c for b 



3) -x = b for x 

5) 3x = ^ for x 

7) E = mc 2 for m 

9) V = g71T 3 for 7T 



11 


a + c = b for c 


13 


c = — — for v 

m + n ^ 


15 


) V = ^ for D 


17 


P = n(p — c) for n 


19 


) T= D ~ d iorD 

Li 


21 


L = L (l + at) for L 


23 


2m + p = 4m + q for m 


25 


k — m r i 

= q tor k 


27 


) h = vt — 16t 2 forv 


29 


) Qi = P(Q2-Qi)iorQ 2 


31 


kA(T+T2) 

a 


33 


ax + b = c for a 


35 


lwh = V for w 


37 


- + b = - for a 

a a 


39 


at — bw = s for t 


41 


ax + bx = c for a 


43 


x + 5y = 3 for y 


45 


3x + 2y = 7 for y 


47 


5a — lb = 4 for b 


49 


4x — 5y = 8 for y 



2) g = - for h 



3.y 



4) p = -f for y 

6 ) z r=l for y 



10 
12 
14 

16 

18 
20 
22 
24 
26 
28 
30 

32 
34 
36 

38 
40 

42 
44 
46 
48 
50 



DS = ds for D 
hj = -—- tor m 

x — f = g for x 
— ^— = k for r 

a — 3 

F = k(i? - L) for k 

S = L + 2B for L 

/ = ^^for£; a 

ax + b = c for x 

q = 6(L — p) for L 

i? = aT + 6 for T 

S 1 = 7rrh + 7rr 2 for h 

L = 7r(ri + r2) + 2d for r*i 



P = 



ViiVi-Vi) 



forVo 



rt = d for r 



V = 
i 



for h 



6 = - for b 

a a 

at — bw = s for w 
x + 5y = 3 for x 
3x + 2y = 7 for x 
5a — 7b = 4 for a 
4x — by = 8 for x 
C = |(F-32)forF 



51 



1.6 

Solving Linear Equations - Absolute Value 

Objective: Solve linear absolute value equations. 

When solving equations with absolute value we can end up with more than one 
possible answer. This is because what is in the absolute value can be either nega- 
tive or positive and we must account for both possibilities when solving equations. 
This is illustrated in the following example. 

Example 84. 

| x | = 7 Absolute value can be positive or negative 
x = 7 or x = — 7 Our Solution 

Notice that we have considered two possibilities, both the positive and negative. 
Either way, the absolute value of our number will be positive 7. 

World View Note: The first set of rules for working with negatives came from 
7th century India. However, in 1758, almost a thousand years later, British math- 
ematician Francis Maseres claimed that negatives "Darken the very whole doc- 
trines of the equations and make dark of the things which are in their nature 
excessively obvious and simple." 

When we have absolute values in our problem it is important to first isolate the 
absolute value, then remove the absolute value by considering both the positive 
and negative solutions. Notice in the next two examples, all the numbers outside 
of the absolute value are moved to the other side first before we remove the abso- 
lute value bars and consider both positive and negative solutions. 

Example 85. 

5 + | x | = 8 Notice absolute value is not alone 

— 5 — 5 Subtract 5 from both sides 

| x | = 3 Absolute value can be positive or negative 
x = 3 or x = — 3 Our Solution 

Example 86. 

— 4\x I = — 20 Notice absolute value is not alone 



Divide both sides by — 4 



52 



\x I = 5 Absolute value can be positive or negative 

x = 5 or x = — 5 Our Solution 



Notice we never combine what is inside the absolute value with what is outside 
the absolute value. This is very important as it will often change the final result 
to an incorrect solution. The next example requires two steps to isolate the abso- 
lute value. The idea is the same as a two-step equation, add or subtract, then 
multiply or divide. 



Example 


87. 












5| 


x\ -4 = 


= 26 








+ 4 +4 








5|x| = 


= 30 








~5~ 


~5~ 








\x\ 


= 6 




X- 


= 6 


or x = 


-6 



Notice the absolute value is not alone 

Add 4 to both sides 

Absolute value still not alone 

Divide both sides by 5 

Absolute value can be positive or negative 

Our Solution 



Again we see the same process, get the absolute value alone first, then consider 
the positive and negative solutions. Often the absolute value will have more than 
just a variable in it. In this case we will have to solve the resulting equations 
when we consider the positive and negative possibilities. This is shown in the next 
example. 



Example 88. 



|2x-l| 
2x - 1 = 7 or 2x - 1 = 



7 Absolute value can be positive or negative 
7 Two equations to solve 



Now notice we have two equations to solve, each equation will give us a different 
solution. Both equations solve like any other two-step equation. 



c-l = 7 




2.x 


-l = -7 


+ 1 + 1 






+ 1 +1 


2x = 8 


or 




2:r=-6 


~2 2" 






~T ~2~ 


x = 4 






x = — 3 



53 



Thus, from our previous example we have two solutions, x = 4 or x ■ 



Again, it is important to remember that the absolute value must be alone first 
before we consider the positive and negative possibilities. This is illustrated in 
below. 



Example 89. 

2 — 4|2ar + 3| = — 18 

To get the absolute value alone we first need to get rid of the 2 by subtracting, 
then divide by — 4. Notice we cannot combine the 2 and — 4 becuase they are 
not like terms, the — 4 has the absolute value connected to it. Also notice we do 
not distribute the — 4 into the absolute value. This is because the numbers out- 
side cannot be combined with the numbers inside the absolute value. Thus we get 
the absolute value alone in the following way: 

2 — 4|2x + 3| = — 18 Notice absolute value is not alone 

— 2 — 2 Subtract 2 from both sides 

— 4|2x + 3 1 = — 20 Absolute value still not alone 

— 4 —4 Divide both sides by — 4 

1 2x + 3 1 =5 Absoloute value can be positive or negative 

2x + 3 = 5 or 2x + 3 = — 5 Two equations to solve 

Now we just solve these two remaining equations to find our solutions. 



x + 3 = 5 




2x + 3 = - 5 


-3-3 




-3 -3 


2x = 2 


or 


2x = -8 


~2~ ~2~ 




~2~ ~2~ 


x= 1 




x=-4 



We now have our two solutions, x = 1 and x = — 4. 



As we are solving absolute value equations it is important to be aware of special 
cases. Remember the result of an absolute value must always be positive. Notice 
what happens in the next example. 

54 



Example 90. 






7+|2a;-5|=4 




-7 -7 



Notice absolute value is not alone 
Subtract 7 from both sides 
\2x — 5 1 = — 3 Result of absolute value is negative! 

Notice the absolute value equals a negative number! This is impossible with abso- 
lute value. When this occurs we say there is no solution or 0. 



One other type of absolute value problem is when two absolute values are equal to 
eachother. We still will consider both the positive and negative result, the differ- 
ence here will be that we will have to distribute a negative into the second abso- 
lute value for the negative possibility. 



Example 91. 

\2x — 7 1 = \Ax + 6 1 Absolute value can be positive or negative 

2x — 7 = 4a; + 6 or 2x — 7 = — (Ax + 6) make second part of second equation negative 

Notice the first equation is the positive possibility and has no significant differ- 
ence other than the missing absolute value bars. The second equation considers 
the negative possibility. For this reason we have a negative in front of the expres- 
sion which will be distributed through the equation on the first step of solving. So 
we solve both these equations as follows: 



2x - 7 = Ax + 6 
- 2x -2x 
- 7 = 2x + 6 
-6 -6 



13 = 2a; 
13 



x 



This gives us our two solutions, x ■ 







2a; -7 = 


= - (Ax + 1 






2a; -7 = 


= - Ax - 6 






+ 4x 


+ 4x 




6a; — 7 = 


= -6 


)T 




+ 7 


+ 7 






6x = 


= 1 






"6" 


"6" 






X- 


_ 1 
= 6 


13 


or 


i 

£ = -. 





55 



1.6 Practice - Absolute Value Equations 



Solve each 


equation. 


1) \x\ = 8 




3) |6| = 1 




5) |5 + 8a| = 


53 


7) |3A; + 8| = 


2 


9) |9 + 7x| = 


30 


11) |8 + 6m 


= 50 


13) |6-2x| 


= 24 


15) — 7| — 3 


— 3r| = — 21 


17) 7|-7x- 


- 3| = 21 


19) | - 46 - 101 


-3 



21) 8|x + 7|-3 = 5 
23) 5|3 + 7m| + l=51 
25) 3 + 5|8-2x|=63 
27) |66 — 2| + 10 = 44 
29) — 7 + 8| — 7x — 3| = 73 
31) |5x + 3| = |2x-l| 
33) |3x-4| = |2x + 3| 



o r \ I 4x - 2 I I 6x + 3 I 

35 ) I^HH^H 



2) \n\ = 7 
4) |x|=2 
6) |9n + 8|=46 
8) |3-x| = 6 
10) |5n + 7|=23 
12) |9p + 6| = 3 
14) |3n-2| = 7 
16) |2 + 26| + 1 = 3 
18) |^M = 2 

20) 8|5p + 8|-5 = ll 
22) 3-|6n + 7| = -40 
24) 4|r + 7| + 3 = 59 
26) 5 + 8|-10n-2| = 101 
28) 7|10v — 2| — 9 = 5 
30) 8|3 — 3n| —5 = 91 
32) |2 + 3x| = |4-2x| 

2x - 5 | I 3x + 4 I 



34) 

36) 



3 I I 2 I 
3x + 2 I I 2x - 3 I 



56 



1.7 

Solving Linear Equations - Variation 

Objective: Solve variation problems by creating variation equations and 
finding the variation constant. 

One application of solving linear equations is variation. Often different events are 
related by what is called the constant of variation. For example, the time it takes 
to travel a certain distance is related to how fast you are traveling. The faster you 
travel, the less time it take to get there. This is one type of variation problem, we 
will look at three types of variation here. Variation problems have two or three 
variables and a constant in them. The constant, usually noted with a k, describes 
the relationship and does not change as the other variables in the problem change. 
There are two ways to set up a variation problem, the first solves for one of the 
variables, a second method is to solve for the constant. Here we will use the 
second method. 

The greek letter pi (it) is used to represent the ratio of the circumference of a 
circle to its diameter. 

World View Note: In the 5th centure, Chinese mathematician Zu Chongzhi cal- 
culated the value of ti to seven decimal places (3.1415926). This was the most 
accurate value of it for the next 1000 years! 

If you take any circle and divide the circumference of the circle by the diameter 
you will always get the same value, about 3.14159... If you have a bigger circum- 
ference you will also have a bigger diameter. This relationship is called direct 
variation or directly proportional. If we see this phrase in the problem we 
know to divide to find the constant of variation. 

Example 92. 

m is varies directly as n "Directly" tells us to divide 

in 

— = k Our formula for the relationship 

n 

In kickboxing, one will find that the longer the board, the easier it is to break. If 
you multiply the force required to break a board by the length of the board you 
will also get a constant. Here, we are multiplying the variables, which means as 
one variable increases, the other variable decreases. This relationship is called 
indirect variation or inversly proportional. If we see this phrase in the 
problem we know to multiply to find the constant of variation. 

Example 93. 

y is inversely proportional to z "Inversely" tells us to multiply 

yz = k Our formula for the relationship 



57 



The formula for the area of a triangle has three variables in it. If we divide the 
area by the base times the height we will also get a constant, -. This relationship 
is called joint variation or jointly proportional. If we see this phrase in the 
problem we know to divide the first variable by the product of the other two to 
find the constant of variation. 

Example 94. 

A varies jointly as x and y "Jointly" tells us to divide by the product 

— = k Our formula for the relationship 
xy 

Once we have our formula for the relationship in a variation problem, we use 
given or known information to calculate the constant of variation. This is shown 
for each type of variation in the next three examples. 

Example 95. 

w is directly proportional to y and w = 50 when y = 5 

w 

— = k 'directly tells us to divide 

. . = k Substitute known values 

(5) 

10 = k Evaluate to find our constant 



Example 96. 

c varies indirectly as d and c = 4. 5 when d = 6 

cd = k "indirectly" tells us to multiply 
(4.5) (6) = k Substitute known values 

27 = k Evaluate to find our constant 

Example 97. 

x is jointly proportional to y and z and x = 48 when y = 2 and z = 4 



x 
— = k "Jointly" tells us to divide by the product 

yz 

' —■ k Substitute known values 



(2) (4) 

6 = k Evaluate to find our constant 



58 



Once we have found the constant of variation we can use it to find other combina- 
tions in the same relationship. Each of these problems we solve will have three 
important steps, none of which should be skipped. 

1. Find the formula for the relationship using the type of variation 

2. Find the constant of variation using known values 

3. Answer the question using the constant of variation 

The next three examples show how this process is worked out for each type of 
variation. 



Example 98. 

The price of an item varies directly with the sales tax. If a $25 item has a sales tax of $2, 
what will the tax be on a $40 item? 

— = k "Directly" tells us to divide price (p) and tax (t) 

(25) 
. . = k Substitute known values for price and tax 

\^) 

12.5 = k Evaluate to find our constant 

40 

— = 12.5 Using our constant, substitute 40 for price to find the tax 



Q^- = 12.5 (t) Multiply by LCD = t to clear fraction 
40 = 12. 5t Reduce the t with the denominator 



12.5 12.5 Divide by 12.5 

3.2 = t Our solution: Tax is $3.20 

Example 99. 

The speed (or rate) Josiah travels to work is inversely proportional to time it 
takes to get there. If he travels 35 miles per hour it will take him 2.5 hours to get 
to work. How long will it take him if he travels 55 miles per hour? 

rt = k "Inversely" tells us to multiply the rate and time 

(35) (2.5) = k Substitute known values for rate and time 

87.5 = A; Evaluate to find our constant 

55t = 87.5 Using our constant, substitute 55 for rate to find the time 

55 55 Divide both sides by 55 

t ~ 1.59 Our solution: It takes him 1.59 hours to get to work 

Example 100. 



59 



The amount of simple interest earned on an investment varies jointly as the prin- 
ciple (amount invested) and the time it is invested. In an account, $150 invested 
for 2 years earned $12 in interest. How much interest would be earned on a $220 
investment for 3 years? 



Pt 

(12) 



k "Jointly" divide Interest (J) byproduct of Principle (P) & time (t) 



= k Substitute known values for Interest , Principle and time 
(150) (2) F 

0.04 = A; Evaluate to find our constant 

0.04 Using constant, substitute 220 for principle and 3 for time 



(220)(3) 
I 



0.04 Evaluate denominator 



660 
' J = 0.04(660) Multiply by 660 to isolate the variable 

I = 26.4 Our Solution: The investment earned $26.40 in interest 

Sometimes a variation problem will ask us to do something to a variable as we set 
up the formula for the relationship. For example, n can be thought of as the ratio 
of the area and the radius squared. This is still direct variation, we say the area 
varies directly as the radius square and thus our variable is squared in our for- 
mula. This is shown in the next example. 

Example 101. 

The area of a circle is directly proportional to the square of the radius. A circle 
with a radius of 10 has an area of 314. What will the area be on a circle of radius 
4? 



r 
(314) 



., — k "Direct" tells us to divide, be sure we use r 2 for the denominator 



(10) 2 
(314) 



A _ 

k Substitute known values into our formula 
k Exponents first 



100 

3.14 = A; Divide to find our constant 

j— ^ = 3.14 Using the constant, use 4 for r , don't forget the squared! 

— = 3.14 Evaluate the exponent 
16 

( 16 ) =3.14(16) Multiply both sides by 16 

,4 = 50.24 Our Solution: Area is 50.24 

When solving variation problems it is important to take the time to clearly state 
the variation formula, find the constant, and solve the final equation. 



60 



1.7 Practice - Variation 

Write the formula that expresses the relationship described 

1. c varies directly as a 

2. x is jointly proportional to y and z 

3. w varies inversely as x 

4. r varies directly as the square of s 

5. f varies jointly as x and y 

6. j is inversely proportional to the cube of m 

7. h is directly proportional to b 

8. x is jointly proportional with the square of a and the square root of b 

9. a is inversely proportional to b 

Find the constant of variation and write the formula to express the 
relationship using that constant 

10. a varies directly as b and a = 15 when b = 5 

11. p is jointly proportional to q and r and p = 12 when q = 8 and r = 3 

12. c varies inversely as d and c = 7 when d = 4 

13. t varies directly as the square of u and t = 6 when u = 3 

14. e varies jointly as f and g and e = 24 when f = 3 and g = 2 

15. w is inversely proportional to the cube of x and w is 54 when x = 3 

16. h is directly proportional to j and h = 12 when j = 8 

17. a is jointly proportional with the square of x and the square root of y and 

a = 25 when x = 5 and y = 9 

18. m is inversely proportional to n and m = 1.8 when n = 2.1 



Solve each of the following variation problems by setting up a formula 
to express the relationship, finding the constant, and then answering 

61 



the question. 

19. The electrical current, in amperes, in a circuit varies directly as the voltage. 
When 15 volts are applied, the current is 5 amperes. What is the current when 

18 volts are applied? 

20. The current in an electrical conductor varies inversely as the resistance of the 
conductor. If the current is 12 ampere when the resistance is 240 ohms, what 
is the current when the resistance is 540 ohms? 

21. Hooke's law states that the distance that a spring is stretched by hanging 
object varies directly as the mass of the object. If the distance is 20 cm when 
the mass is 3 kg, what is the distance when the mass is 5 kg? 

22. The volume of a gas varies inversely as the pressure upon it. The volume of a 
gas is 200 cm 3 under a pressure of 32 kg/cm 2 . What will be its volume under 
a pressure of 40 kg/cm 2 ? 

23. The number of aluminum cans used each year varies directly as the number of 
people using the cans. If 250 people use 60,000 cans in one year, how many 
cans are used each year in Dallas, which has a population of 1,008,000? 

24. The time required to do a job varies inversely as the number of peopel 
working. It takes 5hr for 7 bricklayers to build a park well. How long will it 
take 10 bricklayers to complete the job? 

25. According to Fidelity Investment Vision Magazine, the average weekly 
allowance of children varies directly as their grade level. In a recent year, the 
average allowance of a 9th-grade student was 9.66 dollars per week. What was 
the average weekly allowance of a 4th-grade student? 

26. The wavelength of a radio wave varies inversely as its frequency. A wave with 
a frequency of 1200 kilohertz has a length of 300 meters. What is the length 
of a wave with a frequency of 800 kilohertz? 

27. The number of kilograms of water in a human body varies directly as the 
mass of the body. A 96-kg person contains 64 kg of water. How many kilo 
grams of water are in a 60-kg person? 

28. The time required to drive a fixed distance varies inversely as the speed. It 
takes 5 hr at a speed of 80 km/h to drive a fixed distance. How long will it 
take to drive the same distance at a speed of 70 km/h? 

29. The weight of an object on Mars varies directly as its weight on Earth. A 
person weighs 951b on Earth weighs 38 lb on Mars. How much would a 100-lb 
person weigh on Mars? 

30. At a constant temperature, the volume of a gas varies inversely as the pres- 



62 



sure. If the pressure of a certain gas is 40 newtons per square meter when the 
volume is 600 cubic meters what will the pressure be when the volume is 
reduced by 240 cubic meters? 

31. The time required to empty a tank varies inversely as the rate of pumping. If 
a pump can empty a tank in 45 min at the rate of 600 kL/min, how long will 
it take the pump to empty the same tank at the rate of 1000 kL/min? 

32. The weight of an object varies inversely as the square of the distance from the 
center of the earth. At sea level (6400 km from the center of the earth), an 
astronaut weighs 100 lb. How far above the earth must the astronaut be in 
order to weigh 64 lb? 

33. The stopping distance of a car after the brakes have been applied varies 
directly as the square of the speed r. If a car, traveling 60 mph can stop in 200 

ft, how fast can a car go and still stop in 72 ft? 

34. The drag force on a boat varies jointly as the wetted surface area and the 
square of the velocity of a boat. If a boat going 6.5 mph experiences a drag 
force of 86 N when the wetted surface area is 41.2 ft 2 , how fast must a boat 
with 28.5 ft 2 of wetted surface area go in order to experience a drag force of 

94N? 

35. The intensity of a light from a light bulb varies inversely as the square of the 
distance from the bulb. Suppose intensity is 90 W/m 2 (watts per square 
meter) when the distance is 5 m. How much further would it be to a point 
where the intesity is 40 W/m 2 ? 

36. The volume of a cone varies jointly as its height, and the square of its radius. 
If a cone with a height of 8 centimeters and a radius of 2 centimeters has a 
volume of 33.5 cm 3 , what is the volume of a cone with a height of 6 centime- 
ters and a radius of 4 centimeters? 

37. The intensity of a television signal varies inversely as the square of the dis- 
tance from the transmitter. If the intensity is 25 W/m 2 at a distance of 2 km, 
how far from the trasmitter are you when the intensity is 2.56 W/m 2 ? 

38. The intensity of illumination falling on a surface from a given source of light 
is inversely proportional to the square of the distance from the source of light. 
The unit for measuring the intesity of illumination is usually the footcandle. If 
a given source of light gives an illumination of 1 foot-candle at a distance of 
10 feet, what would the illumination be from the same source at a distance 

of 20 feet? 



63 



1.8 

Linear Equations - Number and Geometry 

Objective: Solve number and geometry problems by creating and 
solving a linear equation. 

Word problems can be tricky. Often it takes a bit of practice to convert the 
English sentence into a mathematical sentence. This is what we will focus on here 
with some basic number problems, geometry problems, and parts problems. 

A few important phrases are described below that can give us clues for how to set 
up a problem. 

• A number (or unknown, a value, etc) often becomes our variable 

• Is (or other forms of is: was, will be, are, etc) often represents equals (=) 
x is 5 becomes x = 5 

• More than often represents addition and is usually built backwards, 
writing the second part plus the first 

Three more than a number becomes x + 3 

• Less than often represents subtraction and is usually built backwards as 
well, writing the second part minus the first 

Four less than a number becomes x — 4 

Using these key phrases we can take a number problem and set up and equation 
and solve. 

Example 102. 

If 28 less than five times a certain number is 232. What is the number? 

5x — 28 Subtraction is built backwards, multiply the unknown by 5 

5x — 28 = 232 Is translates to equals 

+ 28 + 28 Add 28 to both sides 

5a; = 260 The variable is multiplied by 5 

5 5 Divide both sides by 5 

a: = 52 The number is 52. 

This same idea can be extended to a more involved problem as shown in the next 
example. 

Example 103. 

64 





3a; + 15 




6x - 


-10 


3a; + 15 = 


= 6x - 


-10 


-3x 


3x 




15 = 


= 3x - 


-10 


+ 10 


+ 10 




25 = 


= 3a; 




~3~ 


~3~ 




25 
T 


= x 



Fifteen more than three times a number is the same as ten less than six times the 
number. What is the number 

First, addition is built backwards 

Then, subtraction is also built backwards 

Is between the parts tells us they must be equal 

Subtract 3a; so variable is all on one side 

Now we have a two — step equation 

Add 10 to both sides 

The variable is multiplied by 3 

Divide both sides by 3 

Our number is — 

o 

Another type of number problem involves consecutive numbers. Consecutive 
numbers are numbers that come one after the other, such as 3, 4, 5. If we are 
looking for several consecutive numbers it is important to first identify what they 
look like with variables before we set up the equation. This is shown in the fol- 
lowing example. 

Example 104. 

The sum of three consecutive integers is 93. What are the integers? 

First x Make the first number x 

Second x + 1 To get the next number we go up one or + 1 

Third x + 2 Add another 1 (2 total) to get the third 

F + S + T = 93 First (F) plus Second (S) plus Third (T) equals 93 

(x) + (x+ 1) + (x + 2) = 93 Replace F with x , S with x + 1, and T with x + 2 

a; + x + l + x + 2 = 93 Here the parenthesis aren't needed. 

3x + 3 = 93 Combine like terms x + x + x and 2 + 1 

-3-3 Add 3 to both sides 

3x = 90 The variable is multiplied by 3 

3 3 Divide both sides by 3 

x = 30 Our solution for x 

First 30 Replace x in our origional list with 30 

Second (30) + 1 = 31 The numbers are 30, 31, and 32 
Third (30) + 2 = 32 

Sometimes we will work consective even or odd integers, rather than just consecu- 
tive integers. When we had consecutive integers, we only had to add 1 to get to 
the next number so we had x, x + 1, and x + 2 for our first, second, and third 
number respectively. With even or odd numbers they are spaced apart by two. So 
if we want three consecutive even numbers, if the first is x, the next number 
would be x + 2, then finally add two more to get the third, x + 4. The same is 

65 



true for consecutive odd numbers, if the first is x, the next will be x + 2, and the 
third would be x + 4. It is important to note that we are still adding 2 and 4 even 
when the numbers are odd. This is because the phrase "odd" is refering to our x, 
not to what is added to the numbers. Consider the next two examples. 

Example 105. 

The sum of three consecutive even integers is 246. What are the numbers? 

Make the first x 

Even numbers, so we add 2 to get the next 

Add 2 more (4 total) to get the third 

Sum means add First (F) plus Second (S) plus Third (T) 

Replace each F , S , and T with what we labeled them 

Here the parenthesis are not needed 

Combine like terms x + x + x and 2 + 4 

Subtract 6 from both sides 

The variable is multiplied by 3 

Divide both sides by 3 

Our solution for x 

Replace x in the origional list with 80. 

The numbers are 80, 82, and 84. 





First x 




Second a 


: + 2 




Third a 


; + 4 




F+S+T= 


:246 


(x) 


+ (x + 2) + (x + 4) = 


:246 




x+x+2+x+4= 


:246 




Sx + 6 = 


:246 




-6 


-6 




Sx = 


:240 




~3~ 


~3~ 




X 


= 80 




First 80 




Second (80) + 2 : 


= 82 




Third ( 80) + 4 : 


= 84 


Example 106. 





Find three consecutive odd integers so that the sum of twice the first, the second 
and three times the third is 152. 



First x 


Second a 


: + 2 


Third a 


; + 4 


2F + S' + 3T = 


152 


(x) + (a; + 2)+3(x + 4) = 


152 


2x + x + 2 + 3x + 12 = 


152 


6a; + 14 = 


152 


- 14 - 


- 14 


6x = 


138 


~6~ 


~6~ 


x- 


= 23 


First 23 


Second (23) +2: 


= 25 


Third (23) + 4 = 


= 27 



Make the first x 

Odd numbers so we add 2 (same as even!) 

Add 2 more (4 total) to get the third 

Twice the first gives 2F and three times the third gives 3T 

Replace F , S , and T with what we labled them 

Distribute through parenthesis 

Combine like terms 2x + x + Sx and 2 + 14 

Subtract 14 from both sides 

Variable is multiplied by 6 

Divide both sides by 6 

Our solution for x 

Replace x with 23 in the original list 

The numbers are 23, 25, and 27 



66 



When we started with our first, second, and third numbers for both even and odd 
we had x, x + 2, and x + 4. The numbers added do not change with odd or even, 
it is our answer for x that will be odd or even. 

Another example of translating English sentences to mathematical sentences 
comes from geometry. A well known property of triangles is that all three angles 
will always add to 180. For example, the first angle may be 50 degrees, the second 
30 degrees, and the third 100 degrees. If you add these together, 50 + 30 + 100 = 
180. We can use this property to find angles of triangles. 

World View Note: German mathematician Bernhart Thibaut in 1809 tried to 
prove that the angles of a triangle add to 180 without using Euclid's parallel pos- 
tulate (a point of much debate in math history). He created a proof, but it was 
later shown to have an error in the proof. 

Example 107. 

The second angle of a triangle is double the first. The third angle is 40 less than 
the first. Find the three angles. 

First x With nothing given about the first we make that x 

Second 2x The second is double the first , 

Third x - 40 The third is 40 less than the first 

F + S' + T = 180 All three angles add to 180 

(x) + (2x) + (x — 40) = 180 Replace F, S , and T with the labeled values. 

x + 2x + x — 40 = 180 Here the parenthesis are not needed. 

4x — 40 = 180 Combine like terms, x + 2x + x 

+ 40+40 Add 40 to both sides 

4x = 220 The variable is multiplied by 4 

4 4 Divide both sides by 4 

x = 55 Our solution for £ 

First 55 Replace x with 55 in the original list of angles 

Second 2(55) = 110 Our angles are 55, 110, and 15 
Third (55) - 40 = 15 

Another geometry problem involves perimeter or the distance around an object. 
For example, consider a rectangle has a length of 8 and a width of 3. There are 
two lengths and two widths in a rectangle (opposite sides) so we add 8 + 8 + 3 + 
3 = 22. As there are two lengths and two widths in a rectangle an alternative to 
find the perimeter of a rectangle is to use the formula P = 2L + 2W. So for the 
rectangle of length 8 and width 3 the formula would give, P = 2(8) + 2(3) = 16 + 
6 = 22. With problems that we will consider here the formula P = 2L + 2W will be 
used. 

Example 108. 

67 



The perimeter of a rectangle is 44. The length is 5 less than double the width. 
Find the dimensions. 

Length x We will make the length x 

Width 2x — 5 Width is five less than two times the length 

P = 2L + 2W The formula for perimeter of a rectangle 

(44) = 2(x) + 2(2x - 5) Replace P, L, and IT 7 with labeled values 

44 = 2x + 4x — 10 Distribute through parenthesis 

44 = 6x — 10 Combine like terms 2x + 4x 

+ 10 +10 Add 10 to both sides 

54 = 6x The variable is multiplied by 6 

6 6 Divide both sides by 6 

9 = x Our solution for x 

Length 9 Replace x with 9 in the origional list of sides 

Width 2(9) — 5 = 13 The dimensions of the rectangle are 9 by 13. 

We have seen that it is imortant to start by clearly labeling the variables in a 
short list before we begin to solve the problem. This is important in all word 
problems involving variables, not just consective numbers or geometry problems. 
This is shown in the following example. 

Example 109. 

A sofa and a love seat together costs $444. The sofa costs double the love seat. 
How much do they each cost? 

Love Seat x With no information about the love seat, this is our x 

Sofa 2x Sofa is double the love seat , so we multiply by 2 

S + L = 444 Together they cost 444, so we add. 

(x) + (2x) = 444 Replace S and L with labeled values 

3x = 444 Parenthesis are not needed, combine like terms x + 2x 

3 3 Divide both sides by 3 

x = 148 Our solution for x 

Love Seat 148 Replace x with 148 in the origional list 

Sofa 2(148) = 296 The love seat costs $148 and the sofa costs $296. 

Be careful on problems such as these. Many students see the phrase "double" and 
believe that means we only have to divide the 444 by 2 and get $222 for one or 
both of the prices. As you can see this will not work. By clearly labeling the vari- 
ables in the original list we know exactly how to set up and solve these problems. 



68 



1.8 Practice - Number and Geometry Problems 



Solve. 

1. When five is added to three more than a certain number, the result is 19. 
What is the number? 

2. If five is subtracted from three times a certain number, the result is 10. What 
is the number? 

3. When 18 is subtracted from six times a certain number, the result is — 42. 
What is the number? 

4. A certain number added twice to itself equals 96. What is the number? 

5. A number plus itself, plus twice itself, plus 4 times itself, is equal to — 104. 
What is the number? 

6. Sixty more than nine times a number is the same as two less than ten times 
the number. What is the number? 

7. Eleven less than seven times a number is five more than six times the number. 
Find the number. 

8. Fourteen less than eight times a number is three more than four times the 
number. What is the number? 

9. The sum of three consecutive integers is 108. What are the integers? 

10. The sum of three consecutive integers is — 126. What are the integers? 

11. Find three consecutive integers such that the sum of the first, twice the 
second, and three times the third is — 76. 

12. The sum of two consecutive even integers is 106. What are the integers? 

13. The sum of three consecutive odd integers is 189. What are the integers? 



69 



14. The sum of three consecutive odd integers is 255. What are the integers? 

15. Find three consecutive odd integers such that the sum of the first, two times 

the second, and three times the third is 70. 

16. The second angle of a triangle is the same size as the first angle. The third 
angle is 12 degrees larger than the first angle. How large are the angles? 

17. Two angles of a triangle are the same size. The third angle is 12 degrees 
smaller than the first angle. Find the measure the angles. 

18. Two angles of a triangle are the same size. The third angle is 3 times as large 
as the first. How large are the angles? 

19. The third angle of a triangle is the same size as the first. The second angle is 
4 times the third. Find the measure of the angles. 

20. The second angle of a triangle is 3 times as large as the first angle. The third 
angle is 30 degrees more than the first angle. Find the measure of the angles. 

21. The second angle of a triangle is twice as large as the first. The measure of 
the third angle is 20 degrees greater than the first. How large are the angles? 

22. The second angle of a triangle is three times as large as the first. The measure 
of the third angle is 40 degrees greater than that of the first angle. How large 
are the three angles? 

23. The second angle of a triangle is five times as large as the first. The measure 
of the third angle is 12 degrees greater than that of the first angle. How large 
are the angles? 

24. The second angle of a triangle is three times the first, and the third is 12 
degrees less than twice the first. Find the measures of the angles. 

25. The second angle of a triangle is four times the first and the third is 5 degrees 
more than twice the first. Find the measures of the angles. 

26. The perimeter of a rectangle is 150 cm. The length is 15 cm greater than the 
width. Find the dimensions. 

27. The perimeter of a rectangle is 304 cm. The length is 40 cm longer than the 
width. Find the length and width. 

28. The perimeter of a rectangle is 152 meters. The width is 22 meters less than 
the length. Find the length and width. 

29. The perimeter of a rectangle is 280 meters. The width is 26 meters less than 
the length. Find the length and width. 



70 



30. The perimeter of a college basketball court is 96 meters and the length is 14 
meters more than the width. What are the dimensions? 

31. A mountain cabin on 1 acre of land costs $30,000. If the land cost 4 times as 
much as the cabin, what was the cost of each? 

32. A horse and a saddle cost $5000. If the horse cost 4 times as much as the 
saddle, what was the cost of each? 

33. A bicycle and a bicycle helmet cost $240. How much did each cost, if the 
bicycle cost 5 times as much as the helmet? 

34. Of 240 stamps that Harry and his sister collected, Harry collected 3 times as 
many as his sisters. How many did each collect? 

35. If Mr. Brown and his son together had $220, and Mr. Brown had 10 times as 
much as his son, how much money had each? 

36. In a room containing 45 students there were twice as many girls as boys. How 
many of each were there? 

37. Aaron had 7 times as many sheep as Beth, and both together had 608. How 
many sheep had each? 

38. A man bought a cow and a calf for $990, paying 8 times as much for the cow 
as for the calf. What was the cost of each? 

39. Jamal and Moshe began a business with a capital of $7500. If Jamal 
furnished half as much capital as Moshe, how much did each furnish? 

40. A lab technician cuts a 12 inch piece of tubing into two pieces in such a way 
that one piece is 2 times longer than the other. 

41. A 6 ft board is cut into two pieces, one twice as long as the other. How long 
are the pieces? 

42. An eight ft board is cut into two pieces. One piece is 2 ft longer than the 
other. How long are the pieces? 

43. An electrician cuts a 30 ft piece of wire into two pieces. One piece is 2 ft 
longer than the other. How long are the pieces? 

44. The total cost for tuition plus room and board at State University is $2,584. 
Tuition costs $704 more than room and board. What is the tuition fee? 

45. The cost of a private pilot course is $1,275. The flight portion costs $625 
more than the groung school portion. What is the cost of each? 



71 



1.9 



Solving Linear Equations - Age Problems 



Objective: Solve age problems by creating and solving a linear equa- 
tion. 

An application of linear equations is what are called age problems. When we are 
solving age problems we generally will be comparing the age of two people both 
now and in the future (or past). Using the clues given in the problem we will be 
working to find their current age. There can be a lot of information in these prob- 
lems and we can easily get lost in all the information. To help us organize and 
solve our problem we will fill out a three by three table for each problem. An 
example of the basic structure of the table is below 





Age Now 


Change 


Person 1 






Person 2 







Table 6. Structure of Age Table 

Normally where we see "Person 1" and "Person 2" we will use the name of the 
person we are talking about. We will use this table to set up the following 
example. 

Example 110. 

Adam is 20 years younger than Brian. In two years Brian will be twice as old as 
Adam. How old are they now? 





Age Now 


+ 2 


Adam 






Brian 







We use Adam and Brian for our persons 

We use + 2 for change because the second phrase 

is two years in the future 





Age Now 


+ 2 


Adam 


£-20 




Brain 


X 





Consider the "Now" part, Adam is 20 years 
youger than Brian. We are given information about 
Adam, not Brian. So Brian is x now. To show Adam 
is 20 years younger we subtract 20, Adam is x — 20. 





Age Now 


+ 2 


Adam 


z-20 


x - 20 + 2 


Brian 


X 


x + 2 



Now the + 2 column is filled in. This is done by adding 
2 to both Adam's and Brian's now column as shown 
in the table. 





Age Now 


+ 2 


Adam 


x-20 


x-18 


Brian 


x 


x + 2 



Combine like terms in Adam's future age: — 20 + 2 
This table is now filled out and we are ready to try 
and solve. 



72 



B = 2A 

(x + 2) = 2(x - 18) 

x + 2 = 2x-36 
— x — x 



2 = x-36 
+ 36 +36 

38 = x 





Age now 


Adam 


38-20 = 18 


Brian 


38 



Our equation comes from the future statement: 
Brian will be twice as old as Adam. This means 
the younger, Adam, needs to be multiplied by 2. 
Replace B and A with the information in their future 
cells, Adam (A) is replaced with x — 18 and Brian (B) 
is replaced with (x + 2) This is the equation to solve! 

Distribute through parenthesis 

Subtract x from both sides to get variable on one side 

Need to clear the — 36 

Add 36 to both sides 

Our solution for x 

The first column will help us answer the question. 

Replace the x's with 38 and simplify. 

Adam is 18 and Brian is 38 



Solving age problems can be summarized in the following five steps. These five 
steps are guidelines to help organize the problem we are trying to solve. 

1. Fill in the now column. The person we know nothing about is x. 

2. Fill in the future/past collumn by adding/subtracting the change to the 
now column. 

3. Make an equation for the relationship in the future. This is independent of 
the table. 

4. Replace variables in equation with information in future cells of table 

5. Solve the equation for x, use the solution to answer the question 
These five steps can be seen illustrated in the following example. 

Example 111. 

Carmen is 12 years older than David. Five years ago the sum of their ages was 28. 
How old are they now? 





Age Now 


-5 


Carmen 






David 







Five years ago is — 5 in the change column. 





Age Now 


-5 


Carmen 


x + 12 




David 


X 





Carmen is 12 years older than David. We don't 
know about David so he is x , Carmen then is x - 



12 





Age Now 


-5 


Carmen 


a; + 12 


x + 12-5 


David 


X 


x — 5 



Subtract 5 from now column to get the change 



73 





Age Now 


-5 


Carmen 


x + 12 


x + 7 


David 


X 


x — 5 



Simplify by combining like terms 12 
Our table is ready! 



C + D-- 


= 28 


(x + 7) + (x-5) = 


= 28 


x + 7 + x — 5 = 


= 28 


2x + 2 = 


= 28 


-2 


-2 


2a; = 


= 26 


~2~ 


~2~ 


X- 


= 13 



The sum of their ages will be 29. So we add C and D 

Replace C and D with the change cells. 

Remove parenthesis 

Combine like terms x + x and 7 — 5 

Subtract 2 from both sides 

Notice x is multiplied by 2 

Divide both sides by 2 

Our solution for a; 

Replace x with 13 to answer the question 
Carmen is 25 and David is 13 



Sometimes we are given the sum of their ages right now. These problems can be 
tricky. In this case we will write the sum above the now column and make the 
first person's age now x. The second person will then turn into the subtraction 
problem total — x. This is shown in the next example. 

Example 112. 

The sum of the ages of Nicole and Kristin is 32. In two years Nicole will be three 
times as old as Kristin. How old are they now? 





Age Now 


Caremen 


13 + 12 = 25 


David 


13 



32 





Age Now 


+ 2 


Nicole 


X 




Kristen 


32 -x 





The change is + 2 for two years in the future 

The total is placed above Age Now 

The first person is x. The second becomes 32 — x 





Age Now 


+ 2 


Nicole 


X 


x + 2 


Kristen 


32 -x 


32 - x + 2 



Add 2 to each cell fill in the change column 





Age Now 


+ 2 


Nicole 


X 


x + 2 


Kristen 


32 ~x 


34 -x 



Combine like terms 32 + 2, our table is done! 



^ = 3^ 

(x + 2) = 3(34-x) 

x + 2 = 102-3x 

+ 3x +3x 



Nicole is three times as old as Kristin. 

Replace variables with information in change cells 

Distribute through parenthesis 

Add Sx to both sides so variable is only on one side 



74 





4x + 2 = 102 


-2 -2 


4x = 100 


"4" ~4~ 


a; = 25 




Age Now 


Nicole 


25 


Kristen 


32-25 = 7 



Solve the two — step equation 
Subtract 2 from both sides 
The variable is multiplied by 4 
Divide both sides by 4 
Our solution for x 

Plug 25 in for x in the now column 
Nicole is 25 and Kristin is 7 



A slight variation on age problems is to ask not how old the people are, but 
rather ask how long until we have some relationship about their ages. In this case 
we alter our table slightly. In the change column because we don't know the time 
to add or subtract we will use a variable, t, and add or subtract this from the now 
column. This is shown in the next example. 



Example 113. 

Louis is 26 years old. Her daughter is 4 years old. In how many years will Louis 
be double her daughter's age? 



As we are given their ages now, these numbers go into 
the table. The change is unknown, so we write + 1 for 
the change 





Age Now 


+ t 


Louis 


26 




Daughter 


4 







Age Now 


+ t 


Louis 


26 


26 + t 


Daughter 


4 


4 + t 



Fill in the change column by adding t to each person's 
age. Our table is now complete. 





L = 2D 


(26 


+ t) = 2(4 + i) 




26 + t = 8 + 2t 




-t -t 




26 = 8 + t 




-8-8 



18 = t 



Louis will be double her daughter 

Replace variables with information in change cells 

Distribute through parenthesis 

Subtract t from both sides 

Now we have an 8 added to the t 

Subtract 8 from both sides 

In 18 years she will be double her daughter's age 



Age problems have several steps to them. However, if we take the time to work 
through each of the steps carefully, keeping the information organized, the prob- 
lems can be solved quite nicely. 

World View Note: The oldest man in the world was Shigechiyo Izumi from 
Japan who lived to be 120 years, 237 days. However, his exact age has been dis- 
puted. 



75 



1.9 Practice - Age Problems 



1. A boy is 10 years older than his brother. In 4 years he will be twice as old as 
his brother. Find the present age of each. 

2. A father is 4 times as old as his son. In 20 years the father will be twice as old 
as his son. Find the present age of each. 

3. Pat is 20 years older than his son James. In two years Pat will be twice as old 
as James. How old are they now? 

4. Diane is 23 years older than her daughter Amy. In 6 years Diane will be twice 
as old as Amy. How old are they now? 

5. Fred is 4 years older than Barney. Five years ago the sum of their ages was 48. 
How old are they now? 

6. John is four times as old as Martha. Five years ago the sum of their ages was 
50. How old are they now? 

7. Tim is 5 years older than JoAnn. Six years from now the sum of their ages will 
be 79. How old are they now? 

8. Jack is twice as old as Lacy. In three years the sum of their ages will be 54. 
How old are they now? 

9. The sum of the ages of John and Mary is 32. Four years ago, John was twice 
as old as Mary. Find the present age of each. 

10. The sum of the ages of a father and son is 56. Four years ago the father was 3 
times as old as the son. Find the present age of each. 

11. The sum of the ages of a china plate and a glass plate is 16 years. Four years 

ago the china plate was three times the age of the glass plate. Find the 
present age of each plate. 

12. The sum of the ages of a wood plaque and a bronze plaque is 20 years. Four 



76 



years ago, the bronze plaque was one-half the age of the wood plaque. Find 
the present age of each plaque. 

13. A is now 34 years old, and B is 4 years old. In how many years will A be 
twice as old as B? 

14. A man's age is 36 and that of his daughter is 3 years. In how many years will 
the man be 4 times as old as his daughter? 

15. An Oriental rug is 52 years old and a Persian rug is 16 years old. How many 
years ago was the Oriental rug four times as old as the Persian Rug? 

16. A log cabin quilt is 24 years old and a friendship quilt is 6 years old. In how 
may years will the log cabin quilt be three times as old as the friendship 
quilt? 

17. The age of the older of two boys is twice that of the younger; 5 years ago it 
was three times that of the younger. Find the age of each. 

18. A pitcher is 30 years old, and a vase is 22 years old. How many years ago was 
the pitcher twice as old as the vase? 

19. Marge is twice as old as Consuelo. The sum of their ages seven years ago was 
13. How old are they now? 

20. The sum of Jason and Mandy's age is 35. Ten years ago Jason was double 
Mandy's age. How old are they now? 

21. A silver coin is 28 years older than a bronze coin. In 6 years, the silver coin 
will be twice as old as the bronze coin. Find the present age of each coin. 

22. A sofa is 12 years old and a table is 36 years old. In how many years will the 
table be twice as old as the sofa? 

23. A limestone statue is 56 years older than a marble statue. In 12 years, the 
limestone will be three times as old as the marble statue. Find the present age 
of the statues. 

24. A pewter bowl is 8 years old, and a silver bowl is 22 years old. In how many 
years will the silver bowl be twice the age of the pewter bowl? 

25. Brandon is 9 years older than Ronda. In four years the sum of their ages will 
be 91. How old are they now? 

26. A kerosene lamp is 95 years old, and an electric lamp is 55 years old. How 
many years ago was the kerosene lamp twice the age of the electric lamp? 

27. A father is three times as old as his son, and his daughter is 3 years younger 



77 



than the son. If the sum of their ages 3 years ago was 63 years, find the 
present age of the father. 

28. The sum of Clyde and Wendy's age is 64. In four years, Wendy will be three 
times as old as Clyde. How old are they now? 

29. The sum of the ages of two ships is 12 years. Two years ago, the age of the 
older ship was three times the age of the newer ship. Find the present age of 
each ship. 

30. Chelsea's age is double Daniel's age. Eight years ago the sum of their ages 
was 32. How old are they now? 

31. Ann is eighteen years older than her son. One year ago, she was three times 
as old as her son. How old are they now? 

32. The sum of the ages of Kristen and Ben is 32. Four years ago Kristen was 
twice as old as Ben. How old are they both now? 

33. A mosaic is 74 years older than the engraving. Thirty years ago, the mosaic 
was three times as old as the engraving. Find the present age of each. 

34. The sum of the ages of Elli and Dan is 56. Four years ago Elli was 3 times as 

old as Dan. How old are they now? 

35. A wool tapestry is 32 years older than a linen tapestry. Twenty years ago, the 
wool tapestry was twice as old as the linen tapestry. Find the present age of 
each. 

36. Carolyn's age is triple her daughter's age. In eight years the sum of their ages 
will be 72. How old are they now? 

37. Nicole is 26 years old. Emma is 2 years old. In how many years will Nicole be 
triple Emma's age? 

38. The sum of the ages of two children is 16 years. Four years ago, the age of the 
older child was three times the age of the younger child. Find the present age 
of each child. 

39. Mike is 4 years older than Ron. In two years, the sum of their ages will be 84. 
How old are they now? 

40. A marble bust is 25 years old, and a terra-cotta bust is 85 years old. In how 
many years will the terra-cotta bust be three times as old as the marble bust? 



78 



1.10 



Solving Linear Equations - Distance, Rate and Time 



Objective: Solve distance problems by creating and solving a linear 
equation. 

An application of linear equations can be found in distance problems. When 
solving distance problems we will use the relationship rt = d or rate (speed) times 
time equals distance. For example, if a person were to travel 30 mph for 4 hours. 
To find the total distance we would multiply rate times time or (30) (4) = 120. 
This person travel a distance of 120 miles. The problems we will be solving here 
will be a few more steps than described above. So to keep the information in the 
problem organized we will use a table. An example of the basic structure of the 
table is blow: 





Rate 


Time 


Distance 


Person 1 








Person 2 









Table 7. Structure of Distance Problem 



The third column, distance, will always be filled in by multiplying the rate and 
time columns together. If we are given a total distance of both persons or trips we 
will put this information below the distance column. We will now use this table to 
set up and solve the following example 



79 



Example 114. 

Two joggers start from opposite ends of an 8 mile course running towards each 
other. One jogger is running at a rate of 4 mph, and the other is running at a 
rate of 6 mph. After how long will the joggers meet? 





Rate 


Time 


Distance 


Jogger 1 








Jogger 2 









The basic table for the joggers, one and two 





Rate 


Time 


Distance 


Jogger 1 


4 






Jogger 2 


6 







We are given the rates for each jogger. 
These are added to the table 





Rate 


Time 


Distance 


Jogger 1 


4 


t 




Jogger 2 


6 


t 





We only know they both start and end at the 
same time. We use the variable t for both times 





Rate 


Time 


Distance 


Jogger 1 


4 


t 


At 


Jogger 2 


6 


t 


Qt 



8 
At + 6t ■■ 
lOt 



10 10 

4 
5 



t- 



The distance column is filled in by multiplying 
rate by time 

We have total distance, 8 miles, under distance 
The distance column gives equation by adding 
Combine like terms, At + 6t 

Divide both sides by 10 

4 
Our solution for t, — hour (48 minutes) 

o 



As the example illustrates, once the table is filled in, the equation to solve is very 
easy to find. This same process can be seen in the following example 



Example 115. 

Bob and Fred start from the same point and walk in opposite directions. Bob 
walks 2 miles per hour faster than Fred. After 3 hours they are 30 miles apart. 
How fast did each walk? 





Rate 


Time 


Distance 


Bob 




3 




Fred 




3 





The basic table with given times filled in 
Both traveled 3 hours 





Rate 


Time 


Distance 


Bob 


r + 2 


3 




Fred 


r 


3 








Rate 


Time 


Distance 


Bob 


r + 2 


3 


3r + 6 


Fred 


r 


3 


3r 



30 

3r + 6 + 3r = 30 

6r + 6 = 30 

-6-6 

6r = 24 

~6~ ~6~ 

r = 4 





Rate 


Bob 


4 + 2 = 6 


Fred 


4 



Bob walks 2 mph faster than Fred 

We know nothing about Fred, so use r for his rate 

Bob is r + 2, showing 2 mph faster 



Distance column is filled in by multiplying rate by 
Time. Be sure to distribute the 3(r + 2) for Bob. 

Total distance is put under distance 

The distance columns is our equation, by adding 

Combine like terms 3r + 3r 

Subtract 6 from both sides 

The variable is multiplied by 6 

Divide both sides by 6 

Our solution for r 

To answer the question completely we plug 4 in for 

r in the table. Bob traveled 6 miles per hour and 

Fred traveled 4 mph 



Some problems will require us to do a bit of work before we can just fill in the 
cells. One example of this is if we are given a total time, rather than the indi- 
vidual times like we had in the previous example. If we are given total time we 
will write this above the time column, use t for the first person's time, and make 
a subtraction problem, Total — t, for the second person's time. This is shown in 
the next example 



Example 116. 

Two campers left their campsite by canoe and paddled downstream at an average 
speed of 12 mph. They turned around and paddled back upstream at an average 
rate of 4 mph. The total trip took 1 hour. After how much time did the campers 
turn around downstream? 





Rate 


Time 


Distance 


Down 


12 






Up 


4 






1 




Rate 


Time 


Distance 


Down 


12 


t 




Up 


4 


1-t 





Basic table for down and upstream 
Given rates are filled in 



Total time is put above time column 

As we have the total time, in the first time we have 

t , the second time becomes the subtraction, 

total — t 



81 





Rate 


Time 


Distance 


Down 


12 


t 


12t 


Up 


4 


1-t 


4-4* 



12* 


= 4- 


-4* 


+ 4* 


+ 4t 




16* 


= 4 




16 


16 




* = 


1 
"4 



Distance column is found by multiplying rate 
by time. Be sure to distribute 4(1 — *) for 
upstream. As they cover the same distance, 
= is put after the down distance 
With equal sign, distance colum is equation 
Add 4* to both sides so variable is only on one side 
Variable is multiplied by 16 
Divide both sides by 16 

Our solution, turn around after — hr (15 min ) 



Another type of a distance problem where we do some work is when one person 
catches up with another. Here a slower person has a head start and the faster 
person is trying to catch up with him or her and we want to know how long it 
will take the fast person to do this. Our startegy for this problem will be to use * 
for the faster person's time, and add amount of time the head start was to get the 
slower person's time. This is shown in the next example. 



Example 117. 

Mike leaves his house traveling 2 miles per hour. Joy leaves 6 hours later to catch 
up with him traveling 8 miles per hour. How long will it take her to catch up 
with him? 





Rate 


Time 


Distance 


Mike 


2 






Joy 


8 







Basic table for Mike and Joy 
The given rates are filled in 





Rate 


Time 


Distance 


Mike 


2 


t + 6 




Joy 


8 


t 





Joy, the faster person, we use * for time 

Mike's time is * + 6 showing his 6 hour head start 





Rate 


Time 


Distance 


Mike 


2 


t + 6 


2* +12 


Joy 


8 


* 


8* 



2* + 12 = 8* 
It -It 

12 = 6* 
~6~~6~ 



Distance column is found by multiplying the rate 
by time. Be sure to distribute the 2(* + 6) for Mike 
As they cover the same distance, = is put after 
Mike's distance 

Now the distance column is the equation 
Subtract It from both sides 
The variable is multiplied by 6 
Divide both sides by 6 



82 



2 = * Our solution for * , she catches him after 2 hours 



World View Note: The 10,000 race is the longest standard track event. 10,000 
meters is approximately 6.2 miles. The current (at the time of printing) world 
record for this race is held by Ethiopian Kenenisa Bekele with a time of 26 min- 
utes, 17.53 second. That is a rate of 12.7 miles per hour! 

As these example have shown, using the table can help keep all the given informa- 
tion organized, help fill in the cells, and help find the equation we will solve. The 
final example clearly illustrates this. 



Example 118. 

On a 130 mile trip a car travled at an average speed of 55 mph and then reduced 
its speed to 40 mph for the remainder of the trip. The trip took 2.5 hours. For 
how long did the car travel 40 mph? 





Rate 


Time 


Distance 


Fast 


55 






Slow 


40 







Basic table for fast and slow speeds 
The given rates are filled in 



2.5 





Rate 


Time 


Distance 


Fast 


55 


* 




Slow 


40 


2.5 -t 





Total time is put above the time column 
As we have total time, the first time we have * 
The second time is the subtraction problem 
2.5-* 



2.5 





Rate 


Time 


Distance 


Fast 


55 


* 


55* 


Slow 


40 


2.5-* 


100 - 40* 





130 




55* + 100 


-40* = 


130 


15* + 100 = 


130 


— 


100- 


100 




15* = 


= 30 




15 


15 




/ 


= 2 





Time 


Fast 


2 


Slow 


2.5-2 = 0.5 



Distance column is found by multiplying rate 
by time. Be sure to distribute 40(2.5 — *) for slow 

Total distance is put under distance 

The distance column gives our equation by adding 

Combine like terms 55* — 40* 

Subtract 100 from both sides 

The variable is multiplied by 30 

Divide both sides by 15 

Our solution for *. 

To answer the question we plug 2 in for * 

The car traveled 40 mph for 0.5 hours (30 minutes) 



83 



1.10 Practice - Distance, Rate, and Time Problems 



1. A is 60 miles from B. An automobile at A starts for B at the rate of 20 miles 
an hour at the same time that an automobile at B starts for A at the rate of 
25 miles an hour. How long will it be before the automobiles meet? 

2. Two automobiles are 276 miles apart and start at the same time to travel 
toward each other. They travel at rates differing by 5 miles per hour. If they 
meet after 6 hours, find the rate of each. 

3. Two trains travel toward each other from points which are 195 miles apart. 
They travel at rate of 25 and 40 miles an hour respectively. If they start at the 
same time, how soon will they meet? 

4. A and B start toward each other at the same time from points 150 miles apart. 
If A went at the rate of 20 miles an hour, at what rate must B travel if they 
meet in 5 hours? 

5. A passenger and a freight train start toward each other at the same time from 
two points 300 miles apart. If the rate of the passenger train exceeds the rate 
of the freight train by 15 miles per hour, and they meet after 4 hours, what 
must the rate of each be? 

6. Two automobiles started at the same time from a point, but traveled in 
opposite directions. Their rates were 25 and 35 miles per hour respectively. 

After how many hours were they 180 miles apart? 

7. A man having ten hours at his disposal made an excursion, riding out at the 
rate of 10 miles an hour and returning on foot, at the rate of 3 miles an hour. 



84 



Find the distance he rode. 

8. A man walks at the rate of 4 miles per hour. How far can he walk into the 
country and ride back on a trolley that travels at the rate of 20 miles per hour, 
if he must be back home 3 hours from the time he started? 

9. A boy rides away from home in an automobile at the rate of 28 miles an hour 
and walks back at the rate of 4 miles an hour. The round trip requires 2 hours. 
How far does he ride? 

10. A motorboat leaves a harbor and travels at an average speed of 15 mph 
toward an island. The average speed on the return trip was 10 mph. How far 
was the island from the harbor if the total trip took 5 hours? 

11. A family drove to a resort at an average speed of 30 mph and later returned 
over the same road at an average speed of 50 mph. Find the distance to the 
resort if the total driving time was 8 hours. 

12. As part of his flight trainging, a student pilot was required to fly to an airport 
and then return. The average speed to the airport was 90 mph, and the 
average speed returning was 120 mph. Find the distance between the two 
airports if the total flying time was 7 hours. 

13. A, who travels 4 miles an hour starts from a certain place 2 hours in advance 
of B, who travels 5 miles an hour in the same direction. How many hours 
must B travel to overtake A? 

14. A man travels 5 miles an hour. After traveling for 6 hours another man starts 
at the same place, following at the rate of 8 miles an hour. When will the 
second man overtake the first? 

15. A motorboat leaves a harbor and travels at an average speed of 8 mph toward 

a small island. Two hours later a cabin cruiser leaves the same harbor and 
travels at an average speed of 16 mph toward the same island. In how many 
hours after the cabin cruiser leaves will the cabin cuiser be alongside the 
motorboat? 

16. A long distance runner started on a course running at an average speed of 6 

mph. One hour later, a second runner began the same course at an average 
speed of 8 mph. How long after the second runner started will the second 
runner overtake the first runner? 

17. A car traveling at 48 mph overtakes a cyclist who, riding at 12 mph, has had 

a 3 hour head start. How far from the starting point does the car overtake 
the cyclist? 

18. A jet plane traveling at 600 mph overtakes a propeller-driven plane which has 



8o 



had a 2 hour head start. The propeller-driven plane is traveling at 200 mph. 
How far from the starting point does the jet overtake the propeller-driven 
plane? 

19. Two men are traveling in opposite directions at the rate of 20 and 30 miles an 

hour at the same time and from the same place. In how many hours will they 
be 300 miles apart? 

20. Running at an average rate of 8 m/s, a sprinter ran to the end of a track and 

then jogged back to the starting point at an average rate of 3 m/s. The 
sprinter took 55 s to run to the end of the track and jog back. Find the 
length of the track. 

21. A motorboat leaves a harbor and travels at an average speed of 18 mph to an 

island. The average speed on the return trip was 12 mph. How far was the 
island from the harbor if the total trip took 5 h? 

22. A motorboat leaves a harbor and travels at an average speed of 9 mph toward 

a small island. Two hours later a cabin cruiser leaves the same harbor and 
travels at an average speed of 18 mph toward the same island. In how many 
hours after the cabin cruiser leaves will the cabin cruiser be alongside the 
motorboat? 

23. A jet plane traveling at 570 mph overtakes a propeller-driven plane that has 

had a 2 h head start. The propeller- driven plane is traveling at 190 mph. 
How far from the starting point does the jet overtake the propeller-driven 
plane? 

24. Two trains start at the same time from the same place and travel in opposite 

directions. If the rate of one is 6 miles per hour more than the rate of the 
other and they are 168 miles apart at the end of 4 hours, what is the rate of 
each? 

25. As part of flight traning, a student pilot was required to fly to an airport and 

then return. The average speed on the way to the airport was 100 mph, and 
the average speed returning was 150 mph. Find the distance between the two 
airports if the total flight time was 5 h. 

26. Two cyclists start from the same point and ride in opposite directions. One 

cyclist rides twice as fast as the other. In three hours they are 72 miles apart. 
Find the rate of each cyclist. 

27. A car traveling at 56 mph overtakes a cyclist who, riding at 14 mph, has had 

a 3 h head start. How far from the starting point does the car overtake the 
cyclist? 

28. Two small planes start from the same point and fly in opposite directions. 



The first plan is flying 25 mph slower than the second plane. In two hours 
the planes are 430 miles apart. Find the rate of each plane. 

29. A bus traveling at a rate of 60 mph overtakes a car traveling at a rate of 45 

mph. If the car had a 1 h head start, how far from the starting point does 
the bus overtake the car? 

30. Two small planes start from the same point and fly in opposite directions. 

The first plane is flying 25 mph slower than the second plane. In 2 h, the 
planes are 470 mi apart. Find the rate of each plane. 

31. A truck leaves a depot at 11 A.M. and travels at a speed of 45 mph. At noon, 

a van leaves the same place and travels the same route at a speed of 65 mph. 
At what time does the van overtake the truck? 

32. A family drove to a resort at an average speed of 25 mph and later returned 

over the same road at an average speed of 40 mph. Find the distance to the 
resort if the total driving time was 13 h. 

33. Three campers left their campsite by canoe and paddled downstream at an 

average rate of 10 mph. They then turned around and paddled back 
upstream at an average rate of 5 mph to return to their campsite. How long 
did it take the campers to canoe downstream if the total trip took 1 hr? 

34. A motorcycle breaks down and the rider has to walk the rest of the way to 
work. The motorcycle was being driven at 45 mph, and the rider walks at a 
speed of 6 mph. The distance from home to work is 25 miles, and the total 
time for the trip was 2 hours. How far did the motorcycle go before if broke 
down? 

35. A student walks and jogs to college each day. The student averages 5 km/hr 

walking and 9 km/hr jogging. The distance from home to college is 8 km, 
and the student makes the trip in one hour. How far does the student jog? 

36. On a 130 mi trip, a car traveled at an average speed of 55 mph and then 

reduced its speed to 40 mph for the remainder of the trip. The trip took a 
total of 2.5 h. For how long did the car travel at 40 mph? 

37. On a 220 mi trip, a car traveled at an average speed of 50 mph and then 

reduced its average speed to 35 mph for the remainder of the trip. The trip 
took a total of 5 h. How long did the car travel at each speed? 

38. An executive drove from home at an average speed of 40 mph to an airport 

where a helicopter was waiting. The executive boarded the helicopter and 
flew to the corporate offices at and average speed of 60 mph. The entire 
distance was 150 mi. The entire trip took 3 h. Find the distance from the 
airport to the corporate offices. 



87 



Chapter 2 : Graphing 

2.1 Points and Lines 89 

2.2 Slope 95 

2.3 Slope-Intercept Form 102 

2.4 Point-Slope Form 107 

2.5 Parallel and Perpendicular Lines 112 



2.1 



Graphing - Points and Lines 



Objective: Graph points and lines using xy coordinates. 

Often, to get an idea of the behavior of an equation we will make a picture that 
represents the solutions to the equations. A graph is simply a picture of the solu- 
tions to an equation. Before we spend much time on making a visual representa- 
tion of an equation, we first have to understand the basis of graphing. Following 
is an example of what is called the coordinate plane. 

The plane is divided into four sections 
by a horizontal number line (x-axis) 
and a vertical number line (t/-axis). 
Where the two lines meet in the center 
is called the origin. This center origin 
is where x = and y = 0. As we move 
to the right the numbers count up 
from zero, representing # = 1,2,3.... 



To the left the numbers count down from zero, representing x = — 1, — 2, — 3.... 
Similarly, as we move up the number count up from zero, y = 1, 2, 3...., and as we 
move down count down from zero, y = — 1, — 2, — 3. We can put dots on the 
graph which we will call points. Each point has an "address" that defines its loca- 
tion. The first number will be the value on the x — axis or horizontal number line. 
This is the distance the point moves left/right from the origin. The second 
number will represent the value on the y — axis or vertical number line. This is 
the distance the point moves up/down from the origin. The points are given as an 
ordered pair (x,y). 

World View Note: Locations on the globe are given in the same manner, each 
number is a distance from a central point, the origin which is where the prime 
meridian and the equator. This "origin is just off the western coast of Africa. 

The following example finds the address or coordinate pair for each of several 
points on the coordinate plane. 



Example 119. 

Give the coordinates of each point 

. B 



C 



Tracing from the origin, point A is 
right 1, up 4. This becomes A(l, 4). 
Point B is left 5, up 3. Left is back- 
wards or negative so we have B( — 5, 



3). C is straight down 2 units. There 
is no left or right. This means we go 
right zero so the point is C(0, — 2). 



A(1,A),B( - 5, 3), C(0, - 2) Our Solution 



Just as we can give the coordinates for a set of points, we can take a set of points 
and plot them on the plane. 



Example 120. 

Graph the points A(3, 2),B(- 2, 1), C(3, - 4), D( - 2, - 3), E( - 3, 0), F(0, 2), 
G(0,0) 

The first point, A is at (3, 2) this 
means x = 3 (right 3) and y = 2 (up 2). 
Following these instructions, starting 
from the origin, we get our point. 



TTpl 



B 



r 



Left 2 



A 



Up 2 



Right 3 



The second point, B{ - 2, 1), is left 2 
(negative moves backwards), up 1. 
This is also illustrated on the graph. 



Down 



B 



D 



L*eft2 Ri f^3 



JI 



Down 4 



The third point, C(3, - 4) is right 3, 
down 4 (negative moves backwards). 



The fourth point, D ( - 2, - 3) is left 
2, down 3 (both negative, both move 
backwards) 



The last three points have zeros in them. We still treat these points just like the 
other points. If there is a zero there is just no movement. 

Next is E( - 3, 0). This is left 3 (nega- 
tive is backwards), and up zero, right 
on the x — axis. 



E 



Left 3 



D 



Up 



G 



A 



C 



Then is F (0, 2). This is right zero, 
and up two, right on the y — axis. 

Finally is G(0, 0). This point has no 
movement. Thus the point is right on 
the origin. 



90 



E 



B 



D 



F 



G 



A 



Our Solution 



The main purpose of graphs is not to plot random points, but rather to give a 
picture of the solutions to an equation. We may have an equation such as y = 
2x — 3. We may be interested in what type of solution are possible in this equa- 
tion. We can visualize the solution by making a graph of possible x and y combi- 
nations that make this equation a true statement. We will have to start by 
finding possible £ and y combinations. We will do this using a table of values. 



Example 121. 



Graph y = 2x — 3 We make a table of values 



X 


y 


-1 









1 





We will test three values for x. Any three can be used 



X 


y 


-1 


-5 





-3 


1 


-1 



Evaluate each by replacing x with the given value 
x=-l;y = 2(-l)-3 = -2-3 = -5 
^O;^ = 2(0) -3 = 0-3 = -3 
x = l;j/ = 2(l)-3 = 2-3=-l 



( — 1, — 5), (0, — 3), (1, — 1) These then become the points to graph on our equation 




91 



Plot each point. 

Once the point are on the graph, con- 



nect the dots to make a line. 
The graph is our solution 



What this line tells us is that any point on the line will work in the equation y = 
2x — 3. For example, notice the graph also goes through the point (2, 1). If we use 
x = 2, we should get y=l. Sure enough, y = 2(2) — 3 = 4 — 3 = 1, just as the graph 
suggests. Thus we have the line is a picture of all the solutions for y = 2x — 3. We 
can use this table of values method to draw a graph of any linear equation. 



Example 122. 



Graph 2x — 3y = 6 We will use a table of values 



X 


y 


-3 









3 





We will test three values for x. Any three can be used. 



2( — 3) — 3y = 6 Substitute each value in for x and solve for y 

— 6 — 3y = 6 Start with x = — 3, multiply first 
+ 6 +6 Add 6 to both sides 

— 3y = 12 Divide both sides by — 3 
^3 ^3 

y = — 4 Solution for y when x = — 3, add this to table 

2(0)-3y = 6 Nextx = 

— 3y = 6 Multiplying clears the constant term 

— 3 — 3 Divide each side by — 3 

y = — 2 Solution for y when x = 0, add this to table 

2(3)-3y = 6 Next x = 3 

6-3y = 6 Multiply 

— 6 — 6 Subtract 9 from both sides 

— 3^ = Divide each side by — 3 
^3 ^3 

y = Solution for y when x = — 3, add this to table 



92 



X 


y 


-3 


-4 





-2 


3 






Our completed table. 



( — 3, — 4), (0, 2), (3, 0) Table becomes points to graph 




Graph points and connect dots 



Our Solution 



93 



2.1 Practice - Points and Lines 



State the coordinates of each point. 



D 



K 



C 



1)- 



E 



B 



H 



Plot each point. 

2)L(-5,5) K(1,0) J(-3,4) 
I( — 3, 0) H(-4,2) G(4, -2) 
F(-2,-2) E (3,-2) D(0, 3) 
C (0, 4) 

Sketch the graph of each line. 

3 ) V = -jX-3 



5 ) y = -^-4 



7) y = -4x + 2 
9) y = \x-b 
11) y=-^x-3 
13) x + 5y = -15 
15) 4x + y = 5 
17)2x-y = 2 
19) x + y = -l 
21) x -y=-3 



4) y = x-l 
6) y=-^x+l 

8) y = fx + 4 
10) y=-x-2 
12) j/ = ix 
14) 8x-y = 5 
16) 3x + 4t/ = 16 
18) 7x + 3y = -12 
20) 3x + 4y = 8 
22) 9x-y = -4 



94 



2.2 



Graphing - Slope 



Objective: Find the slope of a line given a graph or two points. 

As we graph lines, we will want to be able to identify different properties of the 
lines we graph. One of the most important properties of a line is its slope. Slope 
is a measure of steepness. A line with a large slope, such as 25, is very steep. A 
line with a small slope, such as — is very flat. We will also use slope to describe 
the direction of the line. A line that goes up from left to right will have a positive 
slope and a line that goes down from left to right will have a negative slope. 

As we measure steepness we are interested in how fast the line rises compared to 
how far the line runs. For this reason we will describe slope as the fraction — -. 

*■ run 

Rise would be a vertical change, or a change in the y-values. Run would be a hor- 
izontal change, or a change in the ^-values. So another way to describe slope 
would be the fraction c angemy , R turns out that if we have a graph we can draw 

change in x ° *■ 

vertical and horiztonal lines from one point to another to make what is called a 
slope triangle. The sides of the slope triangle give us our slope. The following 
examples show graphs that we find the slope of using this idea. 



Example 123. 















































































































Ri 


SP 




■4 






































































































Ri 


m 


6 

































































































































To find the slope of this line we will 
consider the rise, or verticle change 
and the run or horizontal change. 
Drawing these lines in makes a slope 
triangle that we can use to count from 
one point to the next the graph goes 
down 4, right 6. This is rise — 4, run 
6. As a fraction it would be, -=-. 

2 6 

Reduce the fraction to get — -. 



Our Solution 



World View Note: When French mathematicians Rene Descartes and Pierre de 
Fermat first developed the coordinate plane and the idea of graphing lines (and 
other functions) the y-axis was not a verticle line! 



Example 124. 



!).-) 























Run 


3 






























































































































Ri 


SP 


fi 































































































































































































To find the slope of this line, the rise 
is up 6, the run is right 3. Our slope is 
then written as a fraction, — or -. 

' run S 

This fraction reduces to 2. This will 
be our slope. 

2 Our Solution 



There are two special lines that have unique slopes that we need to be aware of. 
They are illustrated in the following example. 



Example 125. 



In this graph there is no rise, but the 
run is 3 units. This slope becomes 



0. 



This 



line, 



and all horizontal lines have a zero slope. 



V 



This line has a rise of 5, but no run. 
The slope becomes - = 

undefined. This line, 

and all vertical lines, have no slope. 



As you can see there is a big difference between having a zero slope and having no 
slope or undefined slope. Remember, slope is a measure of steepness. The first 
slope is not steep at all, in fact it is flat. Therefore it has a zero slope. The 
second slope can't get any steeper. It is so steep that there is no number large 
enough to express how steep it is. This is an undefined slope. 

We can find the slope of a line through two points without seeing the points on a 
graph. We can do this using a slope formula. If the rise is the change in y values, 
we can calculate this by subtracting the y values of a point. Similarly, if run is a 
change in the x values, we can calculate this by subtracting the x values of a 
point. In this way we get the following equation for slope. 



The slope of a line through (a?i, j/i) and (a? 2 , 2/2) is 



yi- yi 

x 2 — a?i 



96 



When mathematicians began working with slope, it was called the modular slope. 
For this reason we often represent the slope with the variable m. Now we have 
the following for slope. 



Slope = m = 



rise _ change in y _ j/2 — yi 
run change in x x 2 — x\ 



As we subtract the y values and the x values when calculating slope it is impor- 
tant we subtract them in the same order. This process is shown in the following 
examples. 



Example 126. 



Find the slope between ( — 4,3) and (2, — 9) Identify Xi, 2/1, x 2 , y 2 

(xi, yi) and (x 2 , y 2 ) Use slope formula, m ■ 



m- 



9-3 



m : 



(-4) 
-12 

6 



in- 



2/2-2/1 

X 2 — Xi 



Simplify 

Reduce 
Our Solution 



Example 127. 



Find the slope between (4, 6) and (2,-1) Identify Xi, t/i, x 2 , y 2 

(xi, j/i) and (x 2 , y 2 ) Use slope formula, m 



m 



-1-6 
2-4 



m 



2 
7 



2/2-2/1 

X 2 — X\ 



Simplify 

Reduce, dividing by — 1 



m = — Our Solution 



We may come up against a problem that has a zero slope (horiztonal line) or no 
slope (vertical line) just as with using the graphs. 



Example 128. 



Find the slope between ( — 4, — 1) and ( — 4, — 5) Identify xi, 2/1, x 2 , y 2 



97 



(#i , yi) and (X2, 2/2) Use slope formula, m = 

X2 — X\ 

m = — : 7 7T Simplify 

-4- (-4) 

-4 
m = Can't divide by zero, undefined 

J 

m = no slope Our Solution 

Example 129. 

Find the slope between (3, 1) and ( — 2, 1) Identify xi, y\, X2, 2/2 

(xi, yi) and (X2, 2/2) Use slope formula, m = 

X2 — X\ 

m = — - — - Simplify 

m = Reduce 

— 5 

m = Our Solution 



Again, there is a big difference between no slope and a zero slope. Zero is an 
integer and it has a value, the slope of a flat horizontal line. No slope has no 
value, it is undefined, the slope of a vertical line. 

Using the slope formula we can also find missing points if we know what the slope 
is. This is shown in the following two examples. 



Example 130. 

Find the value of y between the points (2, y) and (5, — 1) with slope — 3 



V2 — Vl 

m = We will plug values into slope formula 

x 2 -x 1 

- 3 = 5 _ 2 ^ Simplify 

- 1- v 

-3 = — — Multiply both sides by 3 

o 

3(3) = — L^!(3) Simplify 

— 9 = — 1 — y Add 1 to both sides 

+ 1 +1 



8 = — y Divide both sides by — 1 
8 = y Our Solution 



98 





m = 


V'2 


- w 




%2 


— X\ 


2 




6- 


2 


5 


X 


-( 


-3) 




2 




4 



Example 131. 

Find the value of x between the points ( — 3, 2) and (x, 6) with slope - 



We will plug values into slope formula 
Simplify 

} Multiply both sides by (x + 3) 

2 

— (x + 3) = 4 Multiply by 5 to clear fraction 

o 

(5)|(x + 3)=4(5) Simplify 

2(x + 3) = 20 Distribute 

2x + 6 = 20 Solve. 

— 6 — 6 Subtract 6 from both sides 

2x = 14 Divide each side by 2 

x = 7 Our Solution 



99 



2.2 Practice - Slope 



Find the slope of each line. 




n 

m 

i 

ii — 



3) 





100 




10) 



Find the slope of the line through each 

12 
14 
16 
18 
20 
22 
24 
26 
28 
30 



11) 


;- 2, 10), (-2, -15) 


13) 


;- 15, 10), (16, -7) 


15) 


10,18), (-11, -10) 


17) 


;- 16, -14), (11, -14) 


19) 


;- 4, 14), (-16, 8) 


21) 


12, -19), (6, 14) 


23) 


;- 5, -10), (-5, 20) 


25) 


;- 17, 19), (10, -7) 


27) 


7,-14), (-8, -9) 


29) 


;- 5, 7), (-18, 14) 



pair of points. 

1,2), (-6,-14) 
13, -2), (7, 7) 
-3, 6), (-20, 13) 
13, 15), (2, 10) 
9,-6), (-7,-7) 
-16,2), (15,-10) 
8, 11), (-3, -13) 
11, -2), (1,17) 
-18, -5), (14, -3) 
19, 15), (5, 11) 



Find the value of x or y so that the line through the points has the 
given slope. 

32) (8, y) and (-2, 4); slope: -\ 

34) (-2, y) and (2, 4); slope: | 
36) (x, — 1) and ( — 4, 6); slope: — — 
38) (2, — 5) and (3, y); slope: 6 
40) (6, 2) and (x, 6); slope: - | 



31) 


[2, 6) and (x, 2); slope: - 


33) 


{ — 3, — 2) and (x, 6); slope: — ■=■ 


35) 


{ — 8,y) and ( — 1,1); slope: - 


37) 


[x, — 7) and ( — 9, — 9); slope: - 


39) 


[x, 5) and (8, 0); slope: — - 



101 



2.3 

Graphing - Slope-Intercept Form 

Objective: Give the equation of a line with a known slope and y-inter- 
cept. 

When graphing a line we found one method we could use is to make a table of 
values. However, if we can identify some properties of the line, we may be able to 
make a graph much quicker and easier. One such method is finding the slope and 
the y-intercept of the equation. The slope can be represented by m and the y- 
intercept, where it crosses the axis and x = 0, can be represented by (0, b) where b 
is the value where the graph crosses the vertical y-axis. Any other point on the 
line can be represented by (x, y). Using this information we will look at the slope 
formula and solve the formula for y. 



Example 132. 



m, (0,6), (x, y) Using the slope formula gives: 
y-b 



x-0 

y-b 



m Simplify 

m Multiply both sides by x 



x 
y — b = mx Add b to both sides 

+ b +b 



y = mx + b Our Solution 

This equation, y = mx + b can be thought of as the equation of any line that as a 
slope of m and a y-intercept of b. This formula is known as the slope-intercept 
equation. 

Slope — Intercept Equation: y = mx + b 

If we know the slope and the y-intercept we can easily find the equation that rep- 
resents the line. 

Example 133. 

3 

Slope = —,y — intercept = — 3 Use the slope — intercept equation 

y = mx + b mis the slope, 6 is the y — intercept 

3 
y = —x — 3 Our Solution 
y 4 

We can also find the equation by looking at a graph and finding the slope and y- 
intercept. 

Example 134. 

102 




Identify the point where the graph 
crosses the y-axis (0,3). This means 
the y- intercept is 3. 

Identity one other point and draw a 
slope triangle to find the slope. The 
slope is — - 



y = mx + . 



Slope-intercept equation 



y 



-x ■ 



Our Solution 



We can also move the opposite direction, using the equation identify the slope 
and y-intercept and graph the equation from this information. However, it will be 
important for the equation to first be in slope intercept form. If it is not, we will 
have to solve it for y so we can identify the slope and the y-intercept. 



Example 135. 






Write in slope - 


- intercept form: 2x — Ay = 6 


Solve for y 




-2x -2x 


Subtract 2x from both sides 




-4y = -2x + 6 


Put x term first 




34 3434 


Divide each term by — 4 




1 3 

V = 77X — — 
y 2 2 


Our Solution 



Once we have an equation in slope-intercept form we can graph it by first plotting 
the y-intercept, then using the slope, find a second point and connecting the dots. 



Example 136. 

Graph y 



m 



2 X - 
y = mx 

1 
"2 



,b- 



^ 



4 Recall the slope — intercept formula 
- b Identity the slope, m, and the y — intercept, b 
4 Make the graph 

Starting with a point at the y-inter- 
cept of — 4, 

Then use the slope — , so we will rise 

*■ run ' 

1 unit and run 2 units to find the next 
point. 

Once we have both points, connect the 
dots to get our graph. 



World View Note: Before our current system of graphing, French Mathemati- 
cian Nicole Oresme, in 1323 sugggested graphing lines that would look more like a 



103 



bar graph with a constant slope! 



Example 


137. 








Graph Sx + 4y = 


= 12 




— 3x 




- 3x 




Ay = - 


-3x + 12 




~4~ 


~T 


4 




y = 


3 


+ 3 




y = 

m = - 


= mx 

4,» 


+ b 
= 3 




Not in slope intercept form 
Subtract 3x from both sides 
Put the x term first 
Divide each term by 4 

Recall slope — intercept equation 
Identity m and b 
Make the graph 



Starting with a point at the y-inter- 
cept of 3, 



Then use the slope — , but its nega- 
j. run cj 

tive so it will go downhill, so we will 

drop 3 units and run 4 units to find 

the next point. 

Once we have both points, connect the 
dots to get our graph. 

We want to be very careful not to confuse using slope to find the next point with 
use a coordinate such as (4, — 2) to find an individule point. Coordinates such as 
(4, — 2) start from the origin and move horizontally first, and vertically second. 
Slope starts from a point on the line that could be anywhere on the graph. The 
numerator is the vertical change and the denominator is the horizontal change. 

Lines with zero slope or no slope can make a problem seem very different. Zero 
slope, or horiztonal line, will simply have a slope of zero which when multiplied 
by x gives zero. So the equation simply becomes y = b or y is equal to the y-coor- 
dinate of the graph. If we have no slope, or a vertical line, the equation can't be 
written in slope intercept at all because the slope is undefined. There is no y in 
these equations. We will simply make x equal to the x-coordinate of the graph. 



Example 138. 



Give the equation of the line in the 
graph. 

Because we have a vertical line and no 
slope there is no slope-intercept equa- 
tion we can use. Rather we make x 
equal to the x-coordinate of — 4 



x - 



4 Our Solution 



104 



2.3 Practice - Slope-Intercept 



Write the slope-intercept form of the equation of each line given the 
slope and the y-intercept. 



1) Slope = 2, y-intercept = 5 

3) Slope = 1, y-intercept = — 4 

5) Slope = — -, y-intercept = — 1 

7) Slope = -, y-intercept = 1 



2) Slope = — 6, y-intercept = 4 
4) Slope = — 1, y-intercept = — 2 
6) Slope = — -, y-intercept = 3 
8) Slope = -, y-intercept = 5 



Write the slope-intercept form of the equation of each line. 

9) 10) 




11 


) 
































12) 






























































































































































































































































































































































































































































































































































































































































































































Vc 








































































14) 



105 




15) x + 10y = -37 
17) 2x + y = -l 
19) 7x-3y = 24 
21) x = -8 
23) y-A = -(x + 5) 
25) y-4 = 4(x-l) 
27) y + 5 = -4(z-2) 
29)y+l = -i(x-4) 



16) x-10y = 3 
18) 6x-lly=-70 
20) Ax + 7y = 28 
22) x-7y=-42 
24) 2/-5 = |(x-2) 

26) 2/ -3 = -|(x + 3) 

28) = x-4 

30) 2 / + 2 = |(a; + 5) 



Sketch the graph of each line. 




31) y = ±x + 4 


32) y = -|x-4 


33) y = ^x-5 


34) y = -\x-\ 


35) y = \x 


36) j/ = - \x + 1 


37) x- ?/ + 3 = 


38) 4x + 5 = 5y 


39) -2/-4 + 3x = 


40) -8 = 6x-2y 


41) -32/ = -5x + 9 


42) -3y = 3-|x 



106 



2.4 

Graphing - Point-Slope Form 

Objective: Give the equation of a line with a known slope and point. 

The slope-intercept form has the advantage of being simple to remember and use, 
however, it has one major disadvantage: we must know the y-intercept in order to 
use it! Generally we do not know the y-intercept, we only know one or more 
points (that are not the y-intercept). In these cases we can't use the slope inter- 
cept equation, so we will use a different more flexible formula. If we let the slope 
of an equation be m, and a specific point on the line be (aji, y\), and any other 
point on the line be (x, y). We can use the slope formula to make a second equa- 
tion. 

Example 139. 

m, (x\, t/i), (x, y) Recall slope formula 
= m Plug in values 

X 2 ~Xi 

= m Multiply both sides by (x — X\) 

y — yi = m(x — xi) Our Solution 

If we know the slope, m of an equation and any point on the line (xi, t/i) we can 
easily plug these values into the equation above which will be called the point- 
slope formula. 

Point — Slope Formula: y — y x = m(x — x±) 

Example 140. 

Write the equation of the line through the point (3, — 4) with a slope of -. 

y — yi = m(x — Xi) Plug values into point — slope formula 
3 
y — ( — 4) = — (x — 3) Simplify signs 
o 
3 
y + 4 = — (x — 3) Our Solution 
o 

Often, we will prefer final answers be written in slope intercept form. If the direc- 

107 



tions ask for the answer in slope-intercept form we will simply distribute the 
slope, then solve for y. 



Example 141. 

Write the equation of the line through the point ( — 6, 2) with a slope of — - in 
slope-intercept form. 



y — t/i = m(x — Xi) Plug values into point — slope formula 

2 
y — 2 = — —{x — ( — 6)) Simplify signs 

2 
y — 2 = — — (x + 6) Distribute slope 
o 
2 
y — 2 = — — x — 4 Solve for y 

+ 2 ' +2 



2 

—x — 2 Our Solution 



An important thing to observe about the point slope formula is that the operation 
between the x's and y's is subtraction. This means when you simplify the signs 
you will have the opposite of the numbers in the point. We need to be very 
careful with signs as we use the point-slope formula. 

In order to find the equation of a line we will always need to know the slope. If 
we don't know the slope to begin with we will have to do some work to find it 
first before we can get an equation. 



Example 142. 

Find the equation of the line through the points ( — 2, 5) and (4, — 3). 



V2 — Vl 

m = First we must find the slope 

X2 — X\ 

m = -. — -T- = — - — = — — Plug values in slope formula and evaluate 

4 — ( — 2) 6 3 

y — t/i = m(x — Xi) With slope and either point, use point — slope formula 
4 
y — 5 = — — (x — ( — 2)) Simplify signs 
o 

4 
y — 5 = — — (x + 2) Our Solution 
o 



Example 143. 



108 



Find the equation of the line through the points ( — 3, 4) and ( — 1, — 2) in slope- 
intercept form. 

V2 — Vl 

m = First we must find the slope 

%2~ X\ 

— 2 — 4 — 6 

m = — -. — — - = — — = — 3 Plug values in slope formula and evaluate 

— 1 — ( — 3) 2 

y — t/i = m(x — Xi) With slope and either point, point — slope formula 

y — 4 = — 3 (a; — ( — 3)) Simplify signs 

y — 4 = — 3(x + 3) Distribute slope 

y — 4 = — 3x — 9 Solve for y 

+ 4 +4 Add 4 to both sides 

y = — Sx — 5 Our Solution 



Example 144. 

Find the equation of the line through the points (6, — 2) and ( — 4, 1) in slope- 
intercept form. 

m = First we must find the slope 

x 2 - Xi 

m = j — -^- = — — = — — Plug values into slope formula and evaluate 

Use slope and either point, use point — slope formula 
Simplify signs 

Distribute slope 

Solve for y. Subtract 2 from both sides 

Using — on right so we have a common denominator 

5 

Our Solution 



World View Note: The city of Konigsberg (now Kaliningrad, Russia) had a 
river that flowed through the city breaking it into several parts. There were 7 
bridges that connected the parts of the city. In 1735 Leonhard Euler considered 
the question of whether it was possible to cross each bridge exactly once and only 
once. It turned out that this problem was impossible, but the work laid the foun- 
dation of what would become graph theory. 







x 2 - 


-Xi 


1- 


-(-2) 


3 


3 




4-6 


-10 


10 




y-yi-- 


= m(x - 


Xi) 


y 


-(-2) = 


3 ( 
1(T 


-6) 




y + 2 = 


3 ( 


-6) 




y + 2 = 


3 

= "To x 


9 

+ 5 
10 




— 2 




5 




y = 


3 

= -To x 


1 
5 



109 



2.4 Practice - Point-Slope Form 



Write the point-slope form of the equation of the line through the 
given point with the given slope. 



I) through (2,3), slope = undefined 
3) through (2, 2), slope = - 

5) through ( — 1,-5), slope = 9 
7) through ( — 4, 1), slope = - 
9) through (0, - 2), slope = - 3 

II) through (0, — 5), slope = — - 

13) through ( — 5, — 3), slope = - 
15) through ( — 1,4), slope = — - 



2) through (1, 2), slope = undefined 
4) through (2, 1), slope = — - 
6) through (2, - 2), slope = - 2 
8) through (4, - 3), slope = - 2 
10) through (- 1,1), slope = 4 
12) through (0,2), slope = - | 
14) through (-1,-4), slope = - | 

Q 

16) through (1, — 4), slope = — - 



Write the slope-intercept form of the equation of the line through the 
given point with the given slope. 



17) through: ( — 1, — 5), slope = 2 

Q 

19) through: (5, — 1), slope = — - 

21) through: ( — 4, 1), slope = - 

23) through: (4, - 2), slope = - | 
25) through: ( — 5, — 3), slope = — - 
27) through: (2, - 2), slope = 1 
29) through: ( — 3,4), slope=undefined 
31) through: ( — 4, 2), slope = — - 



18) through: 
20) through: 
22) through: 

24) through: 
26) through: 
28) through: 
30) through: 
32) through: 



2,-2),slope = -2 
— 2, — 2), slope = — - 
4, -3), slope = -\ 

-2,0), slope = -- 
3, 3), slope = - 
-4, -3), slope = 
-2, -5), slope = 2 
5, 3), slope = \ 



110 



Write the point-slope form of the equation of the line through the 
given points. 



33) through: (-4,3) and (-3,1) 
35) through: (5, 1) and (-3,0) 
37) through: ( - 4, - 2) and (0, 4) 
39) through: (3, 5) and ( — 5,3) 
41) through: (3, - 3) and ( - 4, 5) 



34) through: (1, 3) and ( - 3, 3) 

36) through: (-4,5) and (4, 4) 

38) through: ( - 4, 1) and (4, 4) 

40) through: ( - 1, - 4) and ( - 5, 0) 

42) through: ( - 1, - 5) and ( - 5, - 4) 



Write the slope-intercept form of the equation of the line through the 
given points. 



43) through: (-5,1) and ( - 1, - 2) 
45) through: (-5,5) and (2, - 3) 
47) through: (4, 1) and (1,4) 
49) through: (0,2) and (5, - 3) 
51) through: (0, 3) and ( - 1, - 1) 



44) through: ( - 5, - 1) and (5, - 2) 

46) through: (1,-1) and ( - 5, - 4) 

48) through: (0, 1) and ( - 3, 0) 

50) through: (0, 2) and (2,4) 

52) through: (-2,0) and (5, 3) 



111 



2.5 



Graphing - Parallel and Perpendicular Lines 



Objective: Identify the equation of a line given a parallel or perpendic- 
ular line. 

There is an interesting connection between the slope of lines that are parallel and 
the slope of lines that are perpendicular (meet at a right angle). This is shown in 
the following example. 

Example 145. 




The above graph has 
lines. The slope of 
down 2, run 3, or - 



the 

2 
3" 



two parallel 

top line is 

The slope of 

as 



the bottom line is down 2, run 3 
well, or — -. 




The above graph has two perpendic- 
ular lines. The slope of the flatter line 
is up 2, run 3 or 

-. The slope of the steeper line is down 3, 

3 

run 2 or — -. 



World View Note: Greek Mathematician Euclid lived around 300 BC and pub- 
lished a book titled, The Elements. In it is the famous parallel postulate which 
mathematicians have tried for years to drop from the list of postulates. The 
attempts have failed, yet all the work done has developed new types of geome- 
tries! 

As the above graphs illustrate, parallel lines have the same slope and perpendic- 
ular lines have opposite (one positive, one negative) reciprocal (flipped fraction) 
slopes. We can use these properties to make conclusions about parallel and per- 
pendicular lines. 



Example 146. 

Find the slope of a line parallel to 5y — 2x 



7. 



5y- 


-2x = l 


+ 2x + 2x 


hy~- 


= 2x + l 


~5~ 


T~ T~ 


y = 


2 7 
—x -\ — 
5 5 



To find the slope we will put equation in slope — intercept form 
Add 2x to both sides 
Put x term first 
Divide each term by 5 

The slope is the coefficient of x 



112 



2 
m = — Slope of first line. Parallel lines have the same slope 
5 



m = — Our Solution 

5 

Example 147. 

Find the slope of a line perpendicular to 3x — 4y = 2 

3x — 4y = 2 To find slope we will put equation in slope — intercept form 
— 3x — 3x Subtract 3x from both sides 



— 4y = — 3x + 2 Put x term first 

— 4 — 4—4 Divide each term by — 4 

3 1 
y = — x — — The slope is the coefficient of x 

3 

m = — Slopeoffirstlines. Perpendicular lines have opposite reciprocal slopes 

4 
m = — — Our Solution 
o 

Once we have a slope, it is possible to find the complete equation of the second 
line if we know one point on the second line. 

Example 148. 

Find the equation of a line through (4, — 5) and parallel to 2x — 3y = 6. 

2x — 3y = 6 We first need slope of parallel line 
— 2x —2x Subtract 2x from each side 



3y— — 2x + 6 Put a; term first 



— 3 —3 — 3 Divide each term by — 3 

2 
y = — x — 2 Identify the slope, the coefficient of x 

2 
m = — Parallel lines have the same slope 
o 

2 
m = — We will use this slope and our point (4, — 5) 

y — y± = m(x — x±) Plug this information into point slope formula 

2 
y — ( — 5) = — (x — 4) Simplify signs 
o 

2 
t/ + 5 = — (x — 4) Our Solution 
o 



113 



Example 149. 

Find the equation of the line through (6, — 9) perpendicular to y = — -x + 4 in 
slope-intercept form. 

3 

y = x + 4 Identify the slope, coefficient of x 

5 

3 

m = — — Perpendicular lines have opposite reciprocal slopes 
5 

5 
m = — We will use this slope and our point (6, — 9) 

y — ?/i = m(x — X\) Plug this information into point — slope formula 

5 
y — ( — 9) = —(x — 6) Simplify signs 
o 

5 
y + 9 = — (x — 6) Distribute slope 

5 
y + 9 = —x — 10 Solve for y 
o 

— 9 — 9 Subtract 9 from both sides 



5 
y = — x — 19 Our Solution 
y 3 

Zero slopes and no slopes may seem like opposites (one is a horizontal line, one is 
a vertical line). Because a horizontal line is perpendicular to a vertical line we can 
say that no slope and zero slope are actually perpendicular slopes! 

Example 150. 

Find the equation of the line through (3, 4) perpendicular to x = — 2 

x = — 2 This equation has no slope, o vertical line 

no slope Perpendicular line then would have a zero slope 

m = Use this and our point (3,4) 

y — yi = m(x — Xi) Plug this information into point — slope formula 

y — 4 = 0(x — 3) Distribute slope 

y — 4 = Solve for y 

+ 4 + 4 Add 4 to each side 

y = 4 Our Solution 

Being aware that to be perpendicular to a vertical line means we have a hori- 
zontal line through a y value of 4, thus we could have jumped from this point 
right to the solution, y = 4. 



114 



2.5 Practice - Parallel and Perpendicular Lines 

Find the slope of a line parallel to each given line. 

l)y = 2x + 4 2)y = -|x + 5 

3) y = 4x-5 4) y= _™ x _ 5 

5)x-y = 4 6) 6x -by = 20 

7) 7x + y = - 2 8) 3x + 4y = -8 

Find the slope of a line perpendicular to each given line. 

9) x = S io) y = -\ X -\ 



2 



n ) y=-^x 



4 



3 X I2)y = jx 



13)x-3y = -6 U)3x-y = -3 

15) x + 2y = 8 16) 8x - 3t/ = - 9 

Write the point-slope form of the equation of the line described. 



17 
18 

19 
20 
21 
22 
23 
24 
25 
26 



through: (2, 5) , parallel to x = 
through: (5, 2), parallel to y = -x + 4 

through: (3,4), parallel to y = -x — 5 
through: (1,-1), parallel to y = — -x + 3 
through: (2, 3) , parallel to y = -x + 4 
through: ( — 1,3), parallel to y = — 3x — 1 
through: (4, 2) , parallel to x = 
through: (1,4), parallel to y = -x + 2 
through: (1, — 5), perpendicular to — x + y = 1 
through: (1, — 2), perpendicular to — x + 2y = 2 



115 



27) through: (5,2), perpendicular to 5x + y = — 3 

28) through: (1, 3), perpendicular to — x + y = 1 

29) through: (4, 2) , perpendicular to — 4x + y = 

30) through: ( — 3, — 5), perpendicular to 3x + ly = 

31) through: (2, — 2) perpendicular to 3y — x = 

32) through: ( — 2, 5). perpendicular to y — 2x = 



Write the slope-intercept form of the equation of the line described. 



4, — 3) , parallel to y = — 2x 



5,2), parallel to y = -x 



33; 


through: ( 


34; 


through: ( 


35 


through: ( 


36; 


through: ( 


37 


through: ( 


38 


through: ( 


39; 


through: ( 


40; 


through: ( 


41 


through: ( 


42; 


through: ( 


43; 


through: ( 


44; 


through: ( 


45; 


through: ( 


46; 


through: ( 


47; 


through: ( 


48; 


through: ( 



3, 1), parallel to y = — -x — 1 
4,0), parallel to y = — -x + 4 
4, — 1), parallel to y = — -x+ 1 



2, 3) , parallel to y = -x — 1 
— 2, — 1), parallel to y = — 



-x-2 



— 5,-4), parallel to y = -x — 2 

4, 3) , perpendicular to x + y = — 1 

— 3, — 5), perpendicular to x + 2y = — 4 
5,2), perpendicular to x = 

5, — 1), perpendicular to — 5x + 2y = 10 

— 2, 5), perpendicular to — x + y = — 2 
2, — 3) , perpendicular to — 2x + by = — 10 
4, — 3) , perpendicular to — x + 2y = — 6 

— 4, 1), perpendicular to 4x + 3y = — 9 



116 



Chapter 3 : Inequalities 

3.1 Solve and Graph Inequalities 118 

3.2 Compound Inequalities 124 

3.3 Absolute Value Inequalities 128 



117 



3.1 

Inequalities - Solve and Graph Inequalities 



Objective: Solve, graph, and give interval notation for the solution to 
linear inequalities. 

When we have an equation such as x = 4 we have a specific value for our variable. 
With inequalities we will give a range of values for our variable. To do this we 
will not use equals, but one of the following symbols: 

> Greater than 

^ Greater than or equal to 

< Less than 

^ Less than or equal to 

World View Note: English mathematician Thomas Harriot first used the above 
symbols in 1631. However, they were not immediately accepted as symbols such 
as n and Zl were already coined by another English mathematician, William 
Oughtred. 

If we have an expression such as x < 4, this means our variable can be any number 
smaller than 4 such as — 2, 0, 3, 3.9 or even 3.999999999 as long as it is smaller 



118 



than 4. If we have an expression such as x ^ — 2, this means our variable can be 
any number greater than or equal to — 2, such as 5, 0, — 1, — 1.9999, or even — 2. 

Because we don't have one set value for our variable, it is often useful to draw a 
picture of the solutions to the inequality on a number line. We will start from the 
value in the problem and bold the lower part of the number line if the variable is 
smaller than the number, and bold the upper part of the number line if the vari- 
able is larger. The value itself we will mark with brackets, either ) or ( for less 
than or greater than respectively, and ] or [ for less than or equal to or greater 
than or equal to respectively. 

Once the graph is drawn we can quickly convert the graph into what is called 
interval notation. Interval notation gives two numbers, the first is the smallest 
value, the second is the largest value. If there is no largest value, we can use oo 
(infinity). If there is no smallest value, we can use — oo negative infinity. If we 
use either positive or negative infinity we will always use a curved bracket for that 
value. 



Example 151. 

Graph the inequality and give the interval notation 



x < 2 Start at 2 and shade below 
Use ) for less than 

Our Graph 



-5-4-3-2-1 12 3 4 5 

( — oo , 2) Interval Notation 



Example 152. 

Graph the inequality and give the interval notation 



y ^ — 1 Start at — 1 and shade above 
Use [ for greater than or equal 



" 5 " 4 " 3 -2-10 1 2 3 4 5 Our Graph 

[ — 1 , oo) Interval Notation 

We can also take a graph and find the inequality for it. 



119 



Example 153. 

' ( ID 

Give the inequality for the graph: -5-4-3-2-1 1 2 3* 4 5 
Graph starts at 3 and goes up or greater. Curved bracket means just greater than 

x > 3 Our Solution 



Example 154. 



Give the inequality for the graph: 



<^ 



-> 



-5-4-3-2-10 1 2 3 4 5 



Graph starts at — 4 and goes down or less. Square bracket means less than or 
equal to 

x ^ — 4 Our Solution 

Generally when we are graphing and giving interval notation for an inequality we 
will have to first solve the inequality for our variable. Solving inequalities is very 
similar to solving equations with one exception. Consider the following inequality 
and what happens when various operations are done to it. Notice what happens 
to the inequality sign as we add, subtract, multiply and divide by both positive 
and negative numbers to keep the statment a true statement. 

5>1 Add 3 to both sides 

8 > 4 Subtract 2 from both sides 
6 > 2 Multiply both sides by 3 

12 > 6 Divide both sides by 2 

6>3 Add — 1 to both sides 

5 > 2 Subtract — 4 from both sides 

9 > 6 Multiply both sides by - 2 
— 18< — 12 Divide both sides by -6 

3 > 2 Symbol flipped when we multiply or divide by a negative! 

As the above problem illustrates, we can add, subtract, multiply, or divide on 
both sides of the inequality. But if we multiply or divide by a negative number, 
the symbol will need to flip directions. We will keep that in mind as we solve 
inequalities. 

Example 155. 

Solve and give interval notation 

5 — 2x^11 Subtract 5 from both sides 



120 



-5 

-2a; ^6 

^2~^2 

x < — 3 



00, 



3] 



Divide both sides by — 2 

Divide by a negative — flip symbol! 

Graph, starting at — 3, going down with ] for less than or equal to 

<ii3 - 



-5 -4 -3 -2 
Interval Notation 



-1 1 



The inequality we solve can get as complex as the linear equations we solved. We 
will use all the same patterns to solve these inequalities as we did for solving 
equations. Just remember that any time we multiply or divide by a negative the 
symbol switches directions (multiplying or dividing by a positive does not change 
the symbol!) 



Example 156. 

Solve and give interval notation 



3(2a; - 4) + 4x < 4(3a; - 7) + 8 
6x-12 + 4a;<12x-28 + 8 
10a; - 12 < 12x - 20 
- lOx - 10s 

- 12 < 2x - 20 

+ 20 +20 

8<2x 

~2~ ~2~ 

4<x 



(4,oo) 



Distribute 
Combine like terms 
Move variable to one side 
Subtract 10a; from both sides 
Add 20 to both sides 

Divide both sides by 2 

Be careful with graph, x is larger! 



-5 -4 -3 -2 
Interval Notation 



-1 1 






It is important to be careful when the inequality is written backwards as in the 
previous example (4 < x rather than x > 4). Often students draw their graphs the 
wrong way when this is the case. The inequality symbol opens to the variable, 
this means the variable is greater than 4. So we must shade above the 4. 



121 



3.1 Practice - Solve and Graph Inequalities 

Draw a graph for each inequality and give interval notation. 

1) n> - 5 2) n>4 

3) - 2 ^ k 4) 1 ^ k 

5) 5^x 6) — 5 < x 

Write an inequality for each graph. 

7) 

<i i i 3 - 

-5-4-3-2-10 1 2 3 4 5 
8) 
( I I I I I I 1 I I I l > 



-5-4-3-2-10 1 2 3 4 5 



9) 

< fr 

-5-4-3-2-10 1 2 3 4 5 

10) 

a > 

-5-4-3-2-10 1 2 3 4 5 
ID 



-5-4-3-2-10 1 2 3 4 5 



12) 



-5-4-3-2-10 1 2 3 4 5 



122 



Solve each inequality, graph each solution, and give interval notation. 

14) -2<£ 

16)£<-§ 

18) 11 > 8 + | 
20)^2 

22) ^^-1 

24) -7n- 10^60 

26) 5^| + 1 

28) -8(n-5)^0 

30) -60^-4(-6x-3) 

32) -8(2-2n)^-16 + n 

34) -36 + 6x>-8(x + 2)+4x 

36) 3(ra + 3) + 7(8 - 8n) < 5n + 5 + 2 

38) -(4-5p) + 3^-2(8-5p) 



13) 


i^io 


15) 


2 + r<3 


17) 


8 + |^6 


19) 


2> a-2 
5 


21) 


- 47 ^ 8 - 5x 


23) 


-2(3 + k)<-U 


25) 


18 < - 2( - 8 + p) 


27) 


24>-6(m-6) 


29) 


-r-5(r-6)<-18 


31) 


24 + 46 < 4(1 + 66) 


33) 


-5w-5<-5(4i; + l) 


35) 


4 + 2(a + 5)<-2(-a-4) 


37) 


- (k - 2) > - k - 20 



123 



3.2 

Inequalities - Compound Inequalities 

Objective: Solve, graph and give interval notation to the solution of 
compound inequalities. 

Several inequalities can be combined together to form what are called compound 
inequalities. There are three types of compound inequalities which we will investi- 
gate in this lesson. 

The first type of a compound inequality is an OR inequality. For this type of 
inequality we want a true statment from either one inequality OR the other 
inequality OR both. When we are graphing these type of inequalities we will 
graph each individual inequality above the number line, then move them both 
down together onto the actual number line for our graph that combines them 
together. 

When we give interval notation for our solution, if there are two different parts to 
the graph we will put a U (union) symbol between two sets of interval notation, 
one for each part. 

Example 157. 

Solve each inequality, graph the solution, and give interval notation of solution 

2x — 5 > 3 or 4 — x^6 Solve each inequality 

+ 5 + 5 —4 — 4 Add or subtract first 

2x > 8 or — x ^ 2 Divide 

2 2 — 1 — 1 Dividing by negative flips sign 

x> 4 or x ^ — 2 Graph the inequalities separatly above number line 



e 



-1 1 



** 



( — oo, — 2] U (4, oo) Interval Notation 

World View Note: The symbol for infinity was first used by the Romans, 
although at the time the number was used for 1000. The greeks also used the 
symbol for 10,000. 

There are several different results that could result from an OR statement. The 
graphs could be pointing different directions, as in the graph above, or pointing in 
the same direction as in the graph below on the left, or pointing opposite direc- 
tions, but overlapping as in the graph below on the right. Notice how interval 
notation works for each of these cases. 



124 



<: * <r 



1 



-1 



) I I I I ) ( I I I I I > 

"2345 -5-4-3-2-1012345 



As the graphs overlap, we take the When the graphs are combined they 
largest graph for our solution. cover the entire number line. 

Interval Notation: ( — oo, 1) Interval Notation: ( — oo, oo) or R 

The second type of compound inequality is an AND inequality. AND inequalities 
require both statements to be true. If one is false, they both are false. When we 
graph these inequalities we can follow a similar process, first graph both inequali- 
ties above the number line, but this time only where they overlap will be drawn 
onto the number line for our final graph. When our solution is given in interval 
notation it will be expressed in a manner very similar to single inequalities (there 
is a symbol that can be used for AND, the intersection - fl , but we will not use it 
here). 

Example 158. 

Solve each inequality, graph the solution, and express it interval notation. 

2x + 8 ^ 5x — 7 and 5x — 3 > 3x + 1 Move variables to one side 



8 ^ 3x — 7 and 2x — 3 > 1 Add 7 or 3 to both sides 

7 +7 +3 + 3 



15 ^ 3x and 2x > 4 Divide 
"3" "3" ~2 ~2~ 

5 ^ x and x > 2 Graph, x is smaller (or equal) than 5, 
greater than 2 



e 



3 



* 



2* 3 4 ~ 



-5 -4 -3-2-10 1 2^ 3 4 



(2, 5] Interval Notation 



Again, as we graph AND inequalities, only the overlapping parts of the individual 
graphs makes it to the final number line. As we graph AND inequalities there are 
also three different types of results we could get. The first is shown in the above 



125 



example. The second is if the arrows both point the same way, this is shown 
below on the left. The third is if the arrows point opposite ways but don't 
overlap, this is shown below on the right. Notice how interval notation is 
expressed in each case. 



} 



9 



-1 1 



3 



E: 



\ 



2 3 4 5 -5-4-3-2-101234 






In this graph, the overlap is only the 
smaller graph, so this is what makes it 
to the final number line. 

Interval Notation: ( — oo, — 2) 



In this graph there is no overlap of the 
parts. Because their is no overlap, no 
values make it to the final number 
line. 

Interval Notation: No Solution or 



The third type of compound inequality is a special type of AND inequality. When 
our variable (or expression containing the variable) is between two numbers, we 
can write it as a single math sentence with three parts, such as 5 < x ^ 8, to show 
x is between 5 and 8 (or equal to 8). When solving these type of inequalities, 
because there are three parts to work with, to stay balanced we will do the same 
thing to all three parts (rather than just both sides) to isolate the variable in the 
middle. The graph then is simply the values between the numbers with appro- 
priate brackets on the ends. 

Example 159. 

Solve the inequality, graph the solution, and give interval notation. 



6sC 
■2 



Ax- 



2<2 

2-2 



8 < - Ax < 
"4 34 34 

2^x>0 
Q<x^2 



<r 



Subtract 2 from all three parts 

Divide all three parts by — 4 

Dividing by a negative flips the symbols 

Flip entire statement so values get larger left to right 

Graph x between and 2 



-> 



£+* 



-1 (T 1 



(0, 2] Interval Notation 



126 



3.2 Practice - Compound Inequalities 



Solve each compound inequality, graph its solution, and give interval 
notation. 



1) |^-3or-5n^-10 



3) a; + 7^12or9:r<-45 

5) x — 6 < — 13 or 6x ^ — 60 

7) I > - 1 and v - 2 < 1 

9) -8 + o< - 3 and 4b < 20 

11) a +10^3 and 8a sC 48 

13) 3<9 + x<7 

15) 11 < 8 + k < 12 

17) -3<x-l<l 

19) -4<8-3m<ll 

21) -16sC2n-10^-22 

23) -56 + 10^30and76 + 2^-40 

25) 3x - 9 < 2x + 10 and 5 + 7x < Wx - 10 

27) - 8 - 6v < 8 - 8v and 7v + 9 < 6 + lOv 

29) 1 + 5A; < 7A; - 3 or k - 10 > 2k + 10 

31) 2a; + 9 ^ lOx + 1 and 3x - 2 < 7x + 2 



2) 6m^-24orm-7<-12 

4) 10r>0orr-5<-12 

6) 9 + n<2or5n>40 

8) -9x<63and|<l 

10) -6n<12andJ^2 

12) -6 + O0and2v>4 

14) ^ | ^ - 1 

16) -ll^n-9<-5 

18) K|<0 

20) 3 + Ir > 59 or - 6r - 3 > 33 

22) -6-8x^-6or2 + 10a;>82 

24) n + 10 > 15 or 4n - 5 < - 1 

26) 4n + 8 < 3n - 6 or lOn - 8 ^ 9 + 9n 

28) 5 - 2a ^ 2a + 1 or 10a - 10 ^ 9a + 9 

30) 8-10r^8 + 4ror-6 + 8r<2 + 8r 



32) -9m + 2<-10-6mor-m + 5^10 + 4m 



127 



3.3 

Inequalities - Absolute Value Inequalities 



Objective: Solve, graph and give interval notation for the solution to 
inequalities with absolute values. 

When an inequality has an absolute value we will have to remove the absolute 
value in order to graph the solution or give interval notation. The way we remove 
the absolute value depends on the direction of the inequality symbol. 

Consider \x\ < 2. 

Absolute value is defined as distance from zero. Another way to read this 
inequality would be the distance from zero is less than 2. So on a number line we 
will shade all points that are less than 2 units away from zero. 



<r 



C i i i 3 i ! \ ; 



-1 o 1 



This graph looks just like the graphs of the three part compound inequalities! 
When the absolute value is less than a number we will remove the absolute value 
by changing the problem to a three part inequality, with the negative value on the 
left and the positive value on the right. So \x\ < 2 becomes — 2 < x < 2, as the 
graph above illustrates. 

Consider \x\ > 2. 

Absolute value is defined as distance from zero. Another way to read this 
inequality would be the distance from zero is greater than 2. So on the number 
line we shade all points that are more than 2 units away from zero. 



-1 1 



* 



This graph looks just like the graphs of the OR compound inequalities! When the 
absolute value is greater than a number we will remove the absolute value by 
changing the problem to an OR inequality, the first inequality looking just like 
the problem with no absolute value, the second flipping the inequality symbol and 
changing the value to a negative. So \x \ > 2 becomes x > 2 or x < — 2, as the graph 
above illustrates. 

World View Note: The phrase "absolute value" comes from German mathemati- 
cian Karl Weierstrass in 1876, though he used the absolute value symbol for com- 
plex numbers. The first known use of the symbol for integers comes from a 1939 



128 



edition of a college algebra text! 

For all absolute value inequalities we can also express our answers in interval 
notation which is done the same way it is done for standard compound inequali- 
ties. 

We can solve absolute value inequalities much like we solved absolute value equa- 
tions. Our first step will be to isolate the absolute value. Next we will remove the 
absolute value by making a three part inequality if the absolute value is less than 
a number, or making an OR inequality if the absolute value is greater than a 
number. Then we will solve these inequalites. Remember, if we multiply or divide 
by a negative the inequality symbol will switch directions! 



Example 160. 

Solve, graph, and give interval notation for the solution 

\Ax — 5 1 ^6 Absolute value is greater, use OR 

Ax -5^6 OR 4x - 5 ^ - 6 Solve 

+ 5 + 5 +5 +5 Add 5 to both sides 

4x^11 OR Ax ^ - 1 Divide both sides by 4 
XX XX 

x ^ — - OR x ^ — - Graph 
4 4 



* 



-5-4-3-2-10 1 2 3 4 5 



oo, 



4 



U 



11 

T 



oo 



Interval notation 



Example 161. 

Solve, graph, and give interval notation for the solution 

-4-3|x|^-16 Add 4 to both sides 

+ 4 +4 



3|x|^ — 12 Divide both sides by — 3 
— 3 — 3 Dividing by a negative switches the symbol 
\x\^A Absolute value is greater, use OR 



129 



x^4 OR x^-4 Graph 

*-3 b 

-5-4-3-2-10 1 2 3 ~ 



( — oo, — 4] U [4, oo) Interval Notation 



In the previous example, we cannot combine — 4 and — 3 because they are not like 
terms, the — 3 has an absolute value attached. So we must first clear the — 4 by 
adding 4, then divide by — 3. The next example is similar. 



Example 162. 

Solve, graph, and give interval notation for the solution 

9 — 2|4x + 1| >3 Subtract 9 from both sides 
- 9 - 9 

Divide both sides by — 2 
Dividing by negative switches the symbol 
Absolute value is less, use three part 
Solve 

Subtract 1 from all three parts 
4 < 4x < 2 Divide all three parts by 4 



-2|4x+l|> 


-6 


-2 


-2 


|4x+l| 


<3 


- 3 < 4x + 1 < 3 


-1 -1 


-1 



i r T 
I 

2 



1 < x < — Graph 



-r oi 



-> 



1,-1 Interval Notation 



In the previous example, we cannot distribute the — 2 into the absolute value. 
We can never distribute or combine things outside the absolute value with what is 
inside the absolute value. Our only way to solve is to first isolate the absolute 
value by clearing the values around it, then either make a compound inequality 
(and OR or a three part) to solve. 

130 



It is important to remember as we are solving these equations, the absolute value 
is always positive. If we end up with an absolute value is less than a negative 
number, then we will have no solution because absolute value will always be posi- 
tive, greater than a negative. Similarly, if absolute value is greater than a nega- 
tive, this will always happen. Here the answer will be all real numbers. 



Example 163. 

Solve, graph, and give interval notation for the solution 

12 + 4|6x — 1 1 < 4 Subtract 1 2 from both sides 
-12 -12 



4|6x — 1|< — 8 Divide both sides by 4 



4 4 

1 6a; — 1 1 < — 2 Absolute value can't be less than a negative 



-> 



-5-4-3-2-10 1 2 3 4 5 



No Solution or 



Example 164. 

Solve, graph, and give interval notation for the solution 

5 — 6|x + 7| ^ 17 Subtract 5 from both sides 
-5 -5 



6|x + 7|^12 Divide both sides by — 6 



— 6 — 6 Dividing by a negative flips the symbol 
| x + 7 1 ^ — 2 Absolute value always greater than negative 



-5-4-3-2-10 1 2 3 4 5 



All Real Numbers or R 



131 



3.3 Practice - Absolute Value Inequalities 



Solve each inequality, graph its solution, and give interval notation. 

I) |x|<3 
3) |2x|<6 
5) |x-2|<6 
7) |x-7|<3 
9) |3x — 2| < 9 

II) l + 2|x-l|<9 
13) 6- |2x-5|> = 3 
15) |3x|>5 
17) |x = 3|> = 3 
19) |3rc-5|>>3 
21) 4 + 3|x-l|> = 10 
23) 3 — 2|x — 5| < — 15 
25) —2 — 3|4 — 2x| ^ — 8 
27) 4 — 5| — 2x — 7| < — 1 
29) 3 — 2|4x - 5| ^ 1 
31) —5 — 2|3x — 6| < — 8 
33) 4-4|-2x + 6|>-4 
35) |-10 + x|^8 



2) 


x\^8 




4) 


x + 3|<4 




6) 


r — 8| < 12 




8) 


x + 3|<4 




10) 


|2x + 5|<9 




12) 


10 — 3|x — 2| >4 


14) 


\x\ > 5 




16) 


\x-4 >5 




18) 


|2x-4|>6 




20) 


3- \2-x\ <1 




22) 


3 — 2|3x — 1| > 


-7 


24) 


4 — 6| -6-3x 


<-5 


26) 


— 3 — 2|4ar — 5 


^1 


28) 


-2 + 3|5-x|- 


^4 


30) 


— 2 — 3| — 3x - 


-5^-5 


32) 


6 — 3|1 — 4x| < 


-3 


34) 


— 3 — 4| — 2ar - 


-5|>-7 



132 



Chapter 4 : Systems of Equations 

4.1 Graphing 134 

4.2 Substitution 139 

4.3 Addition/Elimination 146 

4.4 Three Variables 151 

4.5 Application: Value Problems 158 

4.6 Application: Mixture Problems 167 



133 



4.1 

Systems of Equations - Graphing 



Objective: Solve systems of equations by graphing and identifying the 
point of intersection. 

We have solved problems like 3x — 4 = 11 by adding 4 to both sides and then 
dividing by 3 (solution is x = 5). We also have methods to solve equations with 
more than one variable in them. It turns out that to solve for more than one vari- 
able we will need the same number of equations as variables. For example, to 
solve for two variables such as x and y we will need two equations. When we have 
several equations we are using to solve, we call the equations a system of equa- 
tions. When solving a system of equations we are looking for a solution that 
works in both equations. This solution is usually given as an ordered pair (x, y). 
The following example illustrates a solution working in both equations 



Example 165. 

Show (2,1) is the solution to the system _ 

x \~ y — o 

(2,1) Identify x and y from the order d pair 
x = 2, y = 1 Plug these values into each equation 

3(2) — (1) = 5 First equation 

6 — 1 = 5 Evaluate 

5 = 5 True 

(2) + (1) = 3 Second equation, evaluate 
3 = 3 True 



As we found a true statement for both equations we know (2,1) is the solution to 
the system. It is in fact the only combination of numbers that works in both 
equations. In this lesson we will be working to find this point given the equations. 
It seems to follow that if we use points to describe the solution, we can use graphs 
to find the solutions. 

If the graph of a line is a picture of all the solutions, we can graph two lines on 
the same coordinate plane to see the solutions of both equations. We are inter- 



134 



ested in the point that is a solution for both lines, this would be where the lines 
intersect! If we can find the intersection of the lines we have found the solution 
that works in both equations. 



Example 166. 



-x + 3 



y 4 



First: m 



b = 3 



Second: m = — , b- 

4 



To graph we identify slopes and y — intercepts 



Now we can graph both lines on the same plane. 




To graph each equation, we start at 
the y-intercept and use the slope — 

J -L J. run 

to get the next point and connect the 
dots. 

Remember a negative slope is down- 
hill! 

Find the intersection point, (4,1) 

(4,1) Our Solution 



Often our equations won't be in slope-intercept form and we will have to solve 
both equations for y first so we can identity the slope and y-intercept. 



Example 167. 






6x — Sy= — 9 




2x + 2y = -6 


6x — Sy = — 9 


2x + 2y = -6 


— 6x —6x 


-2x -2x 


— 3y = — 6x — 9 


2y = -2x-6 


-3 -3 -3 


2 2 2 


y = 2x + 3 


y = — x — 3 



Solve each equation for y 



Subtract x terms 
Put x terms first 
Divide by coefficient of y 
Identify slope and y — intercepts 



135 



First: m = — , 6 = 3 
Second: m = — — ,b= — 3 



Now we can graph both lines on the same plane 































































































































































































(-?,.- K 









































































































































































To graph each equation, we start at 
the y-intercept and use the slope — 

J -L J. run 

to get the next point and connect the 
dots. 

Remember a negative slope is down- 
hill! 

Find the intersection point, ( — 2, — 1) 

( — 2, — 1) Our Solution 



As we are graphing our lines, it is possible to have one of two unexpected results. 
These are shown and discussed in the next two example. 



Example 168. 



y = - X -4 
3 ■ 1 

y = -x + 1 

3 

First: m = — , b = — 4 
2' 
3 
Second: m = — , b = 1 



Identify slope and y — intercept of each equation 



Now we can graph both equations on the same plane 




To graph each equation, we start at 
the y-intercept and use the slope — 

J L J. run 

to get the next point and connect the 
dots. 

The two lines do not intersect! They 
are parallel! If the lines do not inter- 
sect we know that there is no point 
that works in both equations, there is 
no solution 

No Solution 



We also could have noticed that both lines had the same slope. Remembering 



136 



that parallel lines have the same slope we would have known there was no solu- 
tion even without having to graph the lines. 



Example 169. 





2x- 
Sx - 


6y = 12 
9y = 18 


Solve each equation for y 


2x — 6y = 


= 12 3x- 


-9y = 18 




-2x 


— ■ JU O Jb 


— 3x 


Subtract x terms 


6y = -2x + 12 


-9y=- 


-3x + 18 


Put x terms first 


-6 — 6 — 6 


-9 


-9 -9 


Divide by coefficient of y 


y = ^x-2 


y 


= \x-2 


Identify the slopes and y — intercepts 



First: m = —,b = 
3' 

Second: m = —,b 



Now we can graph both equations together 




To graph each equation, we start at 
the y-intercept and use the slope — 

j s. i run 

to get the next point and connect the 
dots. 

Both equations are the same line! As 
one line is directly on top of the other 
line, we can say that the lines "inter- 
sect" at all the points! Here we say we 
have infinite solutions 



Once we had both equations in slope-intercept form we could have noticed that 
the equations were the same. At this point we could have stated that there are 
infinite solutions without having to go through the work of graphing the equa- 
tions. 

World View Note: The Babylonians were the first to work with systems of 
equations with two variables. However, their work with systems was quickly 
passed by the Greeks who would solve systems of equations with three or four 
variables and around 300 AD, developed methods for solving systems with any 
number of unknowns! 



137 



4.1 Practice - Graphing 

Solve each equation by graphing. 



1) 


y = — x + 1 
y = — hx — 3 


3) 


y=-3 

y = — x — 4 


5) 


3 i i 
y=--X+l 




y=-\x + 2 


7) 
; 


y = \x + 2 

5 A 

j = - -x - 4 


9) 


5 i /i 

t/ = 3X + 4 

2 

2/ = - 3 x - 3 


11 


) X + Sy=-Q 

5x + Sy = 3 


13 


) x — y = 4 
2x + y = -l 


15 


) 2x + Sy = — 6 
2x + y = 2 


17 


) 2x + y = 2 




x — y = 4 


19 


) 2x + y = - 2 
x + 3y = 9 


21 


) = -6x-9y + 36 
12 = 6x-3t/ 


23 


) 2x — y = — 1 
= -2x-t/-3 


25 


) 3 + y = — x 
— 4 — 6x = — y 


27 


— y + 7x = 4 
-y-3 + 7x = 


29 


) -12 + x = 4y 
12-5x = 4y 



2)y=-\x-2 


y = -\x + 2 


4) y = -x -2 
y = -x + 3 


6) y = 2x + 2 


y = — x — 4 


8) y = 2x-4 

y = \x + 2 


10) y = \x + 4 


y = \x + 1 


12) x + 4t/ = -12 
2x + y = 4 


14) 6x + y = -3 
x + y = 2 


16) 3x + 2t/ = 2 
3x + 2y = — 6 


18) x + 2y = 6 
5x — 4y = 16 


20) x-y = 3 

5x + 2y = 8 


22) -2y + x = 4 
2 = -x + ^y 


24) -2y = -4-x 
-2y = -5x + 4 


26) 16 = - x -4y 

-2x = -4-4y 


28) -4 + t/ = x 
x + 2 = — y 


30) -5x + l = -y 
— y + x = — 3 



138 



4.2 



Systems of Equations - Substitution 



Objective: Solve systems of equations using substitution. 

When solving a system by graphing has several limitations. First, it requires the 
graph to be perfectly drawn, if the lines are not straight we may arrive at the 
wrong answer. Second, graphing is not a great method to use if the answer is 
really large, over 100 for example, or if the answer is a decimal the that graph will 
not help us find, 3.2134 for example. For these reasons we will rarely use graphing 
to solve our systems. Instead, an algebraic approach will be used. 

The first algebraic approach is called substitution. We will build the concepts of 
substitution through several example, then end with a five-step process to solve 
problems using this method. 



Example 170. 



x = 5 
y = 2x - 
2/ = 2(5) 
y = 10 

y 



We already know x = 5, substitute this into the other equation 

3 Evaluate, multiply first 

3 Subtract 

7 We now also have y 

(5, 7) Our Solution 



When we know what one variable equals we can plug that value (or expression) in 
for the variable in the other equation. It is very important that when we substi- 
tute, the substituted value goes in parenthesis. The reason for this is shown in the 
next example. 



Example 171. 



2x-3y = 7 



We know y = 3x — 7, substitute this into the other equation 



y = 3x — 7 
2x — 3(3x — 7) = 7 Solve this equation, distributing — 3 first 



139 



2x — 9a; + 21 = 7 Combine like terms 2x — 9x 
-7x + 21 = 7 Subtract 21 
-21-21 



7x = — 14 Divide by — 7 
=T ^7 

x = 2 We now have our x , plug into the y = equation to find y 

= 3(2) — 7 Evaluate, multiply first 

y = 6 — 7 Subtract 

y = — l We now also have y 

(2, — 1) Our Solution 



By using the entire expression 3x — 7 to replace y in the other equation we were 
able to reduce the system to a single linear equation which we can easily solve for 
our first variable. However, the lone variable (a variable without a coefficient) is 
not always alone on one side of the equation. If this happens we can isolate it by 
solving for the lone variable. 



Example 172. 




3x + 2y = 


= 1 


x — by = 


6 


+ hy + hy 


x = Q + by 


3(6 + 5y) + 2y 


= 1 


18 + 15y + 2ty 


= 1 


18 + 17?/ 


= 1 


-18 - 


18 


17?/ = - 


■17 


17 


17 


y = 


-1 


x = 6 + 5(- 


1) 


x = 6 


-5 


X 


= 1 


(1,- 


-1) 



Lone variable is x , isolate by adding 5y to both sides. 

Substitute this into the untouched equation 
Solve this equation, distributing 3 first 
Combine like terms 15y + 2y 
Subtract 18 from both sides 

Divide both sides by 17 

We have our y , plug this into the x = equation to find x 

Evaluate, multiply first 

Subtract 

We now also have x 

Our Solution 



The process in the previous example is how we will solve problems using substitu- 



te 



tion. This process is described and illustrated in the following table which lists 
the five steps to solving by substitution. 



Problem 


4x-2y = 2 
2x + y = — 5 


1. Find the lone variable 


Second Equation, y 

2x + y = -h 


2. Solve for the lone variable 


-2x -2x 
y = — 5 — 2x 


3. Substitute into the untouched equation 


4x-2(-5-2x)=2 


4. Solve 


4x + 10 + 4x = 2 
8^ + 10 = 2 
-10-10 
8x = -8 
~8~ ~8~ 
x = - 1 


5. Plug into lone variable equation and evaluate 


2/=-5-2(-l) 

t/=-5 + 2 
y = -3 


Solution 


(-1,-3) 



Sometimes we have several lone variables in a problem. In this case we will have 
the choice on which lone variable we wish to solve for, either will give the same 
final result. 



Example 


173. 






X 


+ y = 5 






X 


-y = - 


-1 






x + y 


= 5 






-v- 


- y 






x = 5 


-y 


(5- 


-y) 


-y = 


-l 




5- 


-2y = 


-l 




-5 




-5 






-2y = 


-6 






^2 


^2 






y 


= 3 




x = 5 — 


(3) 






X 


= 2 



Find the lone variable: x or y in first, or x in second. 
We will chose x in the first 
Solve for the lone variable, subtract y from both sides 

Plug into the untouched equation, the second equation 
Solve, parenthesis are not needed here, combine like terms 
Subtract 5 from both sides 

Divide both sides by — 2 

We have our y\ 

Plug into lone variable equation, evaluate 

Now we have our x 



141 



(2,3) Our Solution 



Just as with graphing it is possible to have no solution (parallel lines) or 
infinite solutions (same line) with the substitution method. While we won't have 
a parallel line or the same line to look at and conclude if it is one or the other, 
the process takes an interesting turn as shown in the following example. 



Example 174. 



Find the lone variable, y in the first equation 



y + 4 = 3x 
2 y -6x = -8 

y + 4 = 3x Solve for the lone variable, subtract 4 from both sides 
-4 -4 



y = Sx — 4 Plug into untouched equation 

2(3a? — 4) — 6x = — 8 Solve, distribute through parenthesis 

6x — 8 — 6x = — 8 Combine like terms 6x — 6x 

— 8 = — 8 Variables are gone! A true statement. 

Infinite solutions Our Solution 



Because we had a true statement, and no variables, we know that anything that 
works in the first equation, will also work in the second equation. However, we do 
not always end up with a true statement. 



Example 175. 



6x — 3y = — 9 

— 2x + y = 5 

— 2x + y = 5 

+ 2x +2x 

y = 5 + 2x 

6x - 3(5 + 2x) = - 9 

6x — 15 — 6a; = — 9 

-15^-9 

No Solution 



Find the lone variable, y in the second equation 
Solve for the lone variable, add 2x to both sides 

Plug into untouched equation 
Solve, distribute through parenthesis 
Combine like terms 6x — 6x 
Variables are gone! A false statement. 
Our Solution 



Because we had a false statement, and no variables, we know that nothing will 
work in both equations. 



142 



World View Note: French mathematician Rene Descartes wrote a book which 
included an appendix on geometry. It was in this book that he suggested using 
letters from the end of the alphabet for unknown values. This is why often we are 
solving for the variables x, y, and z. 

One more question needs to be considered, what if there is no lone variable? If 
there is no lone variable substitution can still work to solve, we will just have to 
select one variable to solve for and use fractions as we solve. 



Example 176. 



5 5 

28 12y 

5 5 
28(5) 12^/(5) 



5x — 6y = — 14 No lone variable, 

— 2x + 4y = 12 we will solve for x in the first equation 

5x — 6y = — 14 Solve for our variable, add 6y to both sides 
+ 6y +6y 

5x = — 14 + 6y Divide each term by 5 

~5~ ~5~ ~5~ 

- 14 6y 
x = 1 Plug into untouched equation 

5 5 

14 6y\ 

I + Ay = 12 Solve, distribute through parenthesis 



+ 4y = 12 Clear fractions by multiplying by 5 



f- 4y(5) = 12(5) Reduce fractions and multiply 
o 

28 - 12y + 20y = 60 Combine like terms - I2y + 20y 
28 + 8y = 60 Subtract 28 from both sides 
- 28 - 28 



8t/ = 32 Divide both sides by 8 
~8~ "8" 

y = 4 We have our y 

- 14 6(4) 

x = 1 — ^-^ Plug into lone variable equation, multiply 

5 5 

- 14 24 . , , , 

x = + — Add tractions 

5 5 

x = -— Reduce fraction 
5 

x = 2 Now we have our x 
(2,4) Our Solution 



Using the fractions does make the problem a bit more tricky. This is why we have 
another method for solving systems of equations that will be discussed in another 
lesson. 



143 



4.2 Practice - Substitution 



Solve each system by substitution. 

I) y = — 3x 

y = 6x — 9 

3) y = -2x-9 
y = 2x — 1 

5) y = 6x + 4 
y = — 3x — 5 

7) y = 3x + 2 
y= — 3x + 8 

9) y = 2x-3 
y=-2x + 9 

II) y = 6x — 6 
-3x-3t/ = -24 

13) y = -6 

3x — 6y = 30 

15) y=-5 

3x + 4t/ = -17 

17) -2a; + 2y = 18 
y = 7x + 15 

19) t/ = -8x+19 

— x + 6y = 16 

21) 7x-2y = -7 
y = l 

23) x- 5y = 7 
2a: + 7?/ = -20 

25) -2x-j/ = -5 

x-8y = -23 

27) -6x + y = 20 

— 3x — 3y = — 18 

29) 3x + y = 9 
2x + 8y = -16 



2) y = x + 5 
y = - 2x - 4 

4) ?/ = — 6x + 3 
2/ = 6x + 3 

6) y = 3x+ 13 

2/ = -2x-22 

8) y = -2x-9 
y = — 5x — 21 

10) ?/ = 7a; -24 

t/ = — 3x + 16 

12) -x + 3y = 12 
y = 6x + 21 

14) 6x-4y = -8 

y = — 6x + 2 

16) 7x + 2y = -7 
y = 5x + 5 

15) y = x + A 
3x-4t/ = -19 

20) y = -2x + 8 

— 7a: — 6y = — 8 

22) x-2y = -13 

4x + 2y = lS 

2 i) 3x - Ay = 15 
7x + y = 4 

26) 6a: + 4?/ = 16 

- 2x + y = - 3 

28) 7x + 5y = -13 
a; — 4y = — 16 

30) -5x-5y = -20 
-2x + y = 7 



144 



31) 2x + y = 2 

3x + 7y=U 

33) re + 5?/ = 15 

— 3x + 2t/ = 6 

35) -2x + 4t/ = -16 
y = -2 

37) -6rr + 6j/ = -12 
8x — 3y = 16 

39) 2x + 3?/ = 16 

- 7x - y = 20 



32) 2x + y = -7 
5x + 3y = -21 

34) 2x + 3t/ = -10 
7x + y = 3 

36) -2a; + 2?/ = -22 
— 5a: — 7y = — 19 

38) -8x + 2y = -6 
-2x + 3y = ll 

40) -x-4y = -U 
-6x + 8y = 12 



145 



4.3 

Systems of Equations - Addition/Elimination 

Objective: Solve systems of equations using the addition/elimination 
method. 

When solving systems we have found that graphing is very limited when solving 
equations. We then considered a second method known as substituion. This is 
probably the most used idea in solving systems in various areas of algebra. How- 
ever, substitution can get ugly if we don't have a lone variable. This leads us to 
our second method for solving systems of equations. This method is known as 
either Elimination or Addition. We will set up the process in the following exam- 
ples, then define the five step process we can use to solve by elimination. 

Example 177. 

SX ~ Ay = 8 

' ' „, Notice opposites in front of y's. Add columns. 

5x + Ay = - 24 FF y 

8a: = — 16 Solve for x , divide by 8 

"8" ~S~ 

x = — 2 We have our x\ 

5( — 2) + 4y = — 24 Plug into either original equation, simplify 

- 10 + Ay = -2A Add 10 to both sides 

+ 10 +10 



Ay=-1A 


Divide by 4 


~4~ ~A~ 




-7 
V= 2 


Now we have our y\ 


(-2 -J) 


Our Solution 



In the previous example one variable had opposites in front of it, — Ay and Ay. 
Adding these together eliminated the y completely. This allowed us to solve for 
the x. This is the idea behind the addition method. However, generally we won't 
have opposites in front of one of the variables. In this case we will manipulate the 
equations to get the opposites we want by multiplying one or both equations (on 
both sides!). This is shown in the next example. 

Example 178. 

— 6x + 5y = 22 We can get opposites in front of x , by multiplying the 
2x + 3 y = 2 second equation by 3 , to get — 6x and + 6x 

3(2x + Sy) = (2)3 Distribute to get new second equation. 

146 



6x + 9y = 6 New second equation 

— 6x + by = 22 First equation still the same, add 

14y = 28 Divide both sides by 14 
14 14 

t/ = 2 We have our t/! 

2x + 3(2) = 2 Plug into one of the original equations, simplify 

2x + 6 = 2 Subtract 6 from both sides 
-6-6 

2x = — 4 Divide both sides by 2 
X ~2~ 

£ = — 2 We also have our x\ 

(-2,2) Our Solution 

When we looked at the x terms, — 6x and 2x we decided to multiply the 2x by 3 
to get the opposites we were looking for. What we are looking for with our oppo- 
sites is the least common multiple (LCM) of the coefficients. We also could have 
solved the above problem by looking at the terms with y, by and 3y. The LCM of 
3 and 5 is 15. So we would want to multiply both equations, the by by 3, and the 
3y by — 5 to get opposites, lby and — lby. This illustrates an important point, 
some problems we will have to multiply both equations by a constant (on both 
sides) to get the opposites we want. 



We can get opposites in front of x, find LCM of 6 and 9, 
The LCM is 18. We will multiply to get 18y and - 18y 



Example 179. 




Sx + 6y = - 


9 


2x + 9y = - 


26 


3(3x + 6y) = (- 


9)3 


9x + l8y=- 


-27 


-2(2x + 9y) = (-26)(- 


-2) 


-Ax- 18y~- 


= 52 


9x + 18y=- 


-27 


-4x- 18y~- 


= 52 


bx - 


= 25 


T~ 


T~ 


X 


= 5 


3(5) + 6y = 


-9 


15 + 6y = 


-9 


-15 


-15 



Multiply the first equation by 3, both sides! 

Multiply the second equation by — 2, both sides! 

Add two new equations together 

Divide both sides by 5 

We have our solution for x 

Plug into either original equation, simplify 

Subtract 15 from both sides 



147 



6y= — 24 Divide both sides by 6 

y = — 4 Now we have our solution for y 
(5, — 4) Our Solution 



It is important for each problem as we get started that all variables and constants 
are lined up before we start multiplying and adding equations. This is illustrated 
in the next example which includes the five steps we will go through to solve a 
problem using elimination. 



Problem 


2x-5y = -13 
— 3y + 4 = — 5a; 


1. Line up the variables and constants 


Second Equation: 
-3y + 4 = -5a; 
+ 5x - 4 + 5x - 4 
5a; — Sy = — 4 


2. Multiply to get opposites (use LCD) 


2x-5y=-13 

5a; — 3y = — 4 

First Equation: multiply by — 5 
-5(2x-5t/) = (-13)(-5) 
-10x + 25^ = 65 

Second Equation: multiply by 2 

2(5x-3t/) = (-4)2 

lO X -Qy = -8 

-10x + 25^ = 65 
10x-6y = -S 


3. Add 


19y = 57 


4. Solve 


19y = 57 

T9" T9" 
y = 3 


5. Plug into either original and solve 


2x -5(3) = -13 
2a; -15 = -13 
+ 15 +15 
2a; =2 
~2~ ~T 
x = l 


Solution 


(1,3) 



World View Note: The famous mathematical text, The Nine Chapters on the 
Mathematical Art, which was printed around 179 AD in China describes a for- 
mula very similar to Gaussian elimination which is very similar to the addition 
method. 



148 



Just as with graphing and substution, it is possible to have no solution or infinite 
solutions with elimination. Just as with substitution, if the variables all disappear 
from our problem, a true statment will indicate infinite solutions and a false stat- 
ment will indicate no solution. 

Example 180. 



2x — 5y = 3 
- 6x + 15?/ = — 9 



To get opposites in front of x , multiply first equation by 3 



3(2x - by) = (3)3 Distribute 
6x — I5y = 9 

6x — 15y = 9 Add equations together 
— 6x + 15ty = — 9 



= True statement 
Infinite solutions Our Solution 



Example 181. 



4x-6y = 8 
6x — 9y = 15 



LCMforx'sisl2. 



3(4x — 6y) = (8)3 Multiply first equation by 3 
12x-18t/ = 24 

2(6x — 9y) = (15) ( — 2) Multiply second equation by — 2 
-12x + 18y = -30 

12a; — 18y = 24 Add both new equations together 
-12x + 18y = -30 



= — 6 False statement 
No Solution Our Solution 

We have covered three different methods that can be used to solve a system of 
two equations with two variables. While all three can be used to solve any 
system, graphing works great for small integer solutions. Substitution works great 
when we have a lone variable, and addition works great when the other two 
methods fail. As each method has its own strengths, it is important you are 
familiar with all three methods. 



149 



4.3 Practice - Addition /Elimination 



Solve each system by elimination. 




1) 4x + 2y = 


2) -7x + ?/ = -10 


-4x-9y = -28 


_ 9x-y = -22 


3) -9x + 5y = -22 


4) -x-2y = -7 


9x — 5y= 13 


x + 2y = 7 


5) -6x + 9y = 3 


6) 5x — 5y = — 15 


6x — 9y= — 9 


5x — 5t/= — 15 


7) 4x-6y = -10 


8) -3x + 3y = -12 


4x — 6y = — 14 


-3x + 9y = -24 


9) -x-5y = 28 


10) -10x-5y = 


- x + Ay = - 17 


- lOrr - 10y = - 30 


11) 2x-y = h 


12) -5a: + 6y = -17 


/ if 

5x + 2y = -28 


x — 2y = 5 


13) 10x + 6t/ = 24 
— 6x + y = 4 


14) x + 3y = -l 
10x + 6y = -10 




16) -6a: + 4y=12 


15) 2x + 4y = 24 
4a: -12?/ = 8 


12x + 6y = 18 




18) -6x + 4y = 4 


17) -7x + 4y = -4 


— 3x — y = 26 


lO X -gy=-g 






20) -9x-5y = -19 


19) 5a: + 10?/ = 20 


3x — 7y = — 11 


— 6x — by = — 3 






22) -5x + 4?/ = 4 


21) -7x-3y = 12 


-7a: -10?/ = -10 


— 6x — 5?/ = 20 


24) 3x + 7y = - 8 


23) 9rr-2j/ = -18 


4x + 6y = — 4 


5a; — 7y = — 10 


26) -4x-5y=12 


25) 9x + 6y = -21 


-10x + 6y = 30 


-10x-9y = 28 


28) 8x + 7y = -24 


27) -7a: + 5t/ = -8 


6x + 3t/ = -18 


-3x-3t/ = 12 


30) -7x + 10y = 13 


29) -8x-8y = -8 


4a: + 9^ = 22 


10a: + 9y = l 


32) = -9x-21 + 12t/ 


31) 9y = 7-x 


1 + 3 2/+ 3^=0 


- 18y + 4a: = - 26 


34) -6-42?/ = - 12a: 


33) = 9x + 5y 


1 7 n 


y = Y x 





150 



4.4 

Systems of Equations - Three Variables 



Objective: Solve systems of equations with three variables using addi- 
tion/elimination. 

Solving systems of equations with 3 variables is very similar to how we solve sys- 
tems with two variables. When we had two variables we reduced the system down 
to one with only one variable (by substitution or addition). With three variables 
we will reduce the system down to one with two variables (usually by addition), 
which we can then solve by either addition or substitution. 

To reduce from three variables down to two it is very important to keep the work 
organized. We will use addition with two equations to eliminate one variable. 
This new equation we will call (A). Then we will use a different pair of equations 
and use addition to eliminate the same variable. This second new equation we 
will call (B). Once we have done this we will have two equations (A) and (B) 
with the same two variables that we can solve using either method. This is shown 
in the following examples. 



Example 182. 

Sx + 2y — z = — 1 
— 2x — 2y + 3z = 5 We will eliminate y using two different pairs of equations 
5x + 2y — z = S 



151 



3x + 1y — z = — 1 Using the first two equations, 
- 2x — 2y + 3z = 5 Add the first two equations 



(A) x + 2z = 4 This is equation (A) , our first equation 

— 2x — 2y + 3z = 5 Using the second two equations 

5x + 2y — z = 3 Add the second two equations 

(B) 3x +2z = 8 This is equation (B), our second equation 

(A) x + 2z = 4 Using (A) and (B) we will solve this system. 

(B) Sx + 2z = 8 We will solve by addition 

-l(x + 2z) = (4)(-l) Multiply (A) by - 1 
-x-2z=-A 

— x — 2z = — 4 Add to the second equation, unchanged 
3x + 2^ = 8 

2x = 4 Solve, divide by 2 
~T ~2 

x = 2 We now have x\ Plug this into either (A) or (B) 

(2) + 2z = 4 We plug it into (A) , solve this equation, subtract 2 
-2 -2 



2z = 2 Divide by 2 
~2" ^~ 
2 = 1 We now have z\ Plug this and x into any original equation 

3(2) + 2y — (1) = — 1 We use the first, multiply 3(2) = 6 and combine with — 1 

2y + 5 = — 1 Solve, subtract 5 
-5 -5 

2y=-6 Divide by 2 
~2 ~2 

y = — 3 We now have y\ 

(2,-3,1) Our Solution 

As we are solving for x, y, and z we will have an ordered triplet (x,y, z) instead of 

152 



just the ordered pair (x, y). In this above problem, y was easily eliminated using 
the addition method. However, sometimes we may have to do a bit of work to get 
a variable to eliminate. Just as with addition of two equations, we may have to 
multiply equations by something on both sides to get the opposites we want so a 
variable eliminates. As we do this remmeber it is improtant to eliminate the 
same variable both times using two different pairs of equations. 



Example 183. 

4x — 3y + 2z = — 29 No variable will easily eliminate. 

6x + 2y — z = — 16 We could choose any variable, so we chose x 

— 8x — y + 3z = 23 We will eliminate x twice. 

4x — 3y + 2z = — 29 Start with first two equations. LCMof4and6is 12. 

6x + 2y — z = — 16 Make the first equation have 12x, the second — \2x 

3(4x -3y + 2z) = (- 29)3 Multiply the first equation by 3 
12x-9y + 6z = -87 

— 2(6:r + 2y — z) = ( — 16) ( — 2) Multiply the second equation by — 2 

- \2x - 4y + 2z = 32 

12x — 9y + 6z = — 87 Add these two equations together 
-12x-4y + 2z = 32 

(A) — 13y + 8z = — 55 This is our (A) equation 

6x + 2y — z = — 16 Now use the second two equations (a different pair) 

-8x-y + 3z = 23 The LCM of 6 and - 8 is 24. 

4(6x + 2y — z) = ( — 16)4 Multiply the first equation by 4 
24x + 8y - 4 = - 64 

3( — 8x — y + 3z) = (23)3 Multiply the second equation by 3 
-24x-3y + 9z = 69 

24x + 8y — 4 = — 64 Add these two equations together 

- 24x - 3y + 9z = 69 



(B) by + 5z = 5 This is our (£>) equation 



153 



(A) — 13y + 8z = — 55 Using (A) and (B) we will solve this system 

(B) 5y + 5z = 5 The second equation is solved for z to use substitution 

5y + 5z = 5 Solving for z , subtract 5y 
-5y -5y 

5z = 5 — 5y Divide each term by 5 
T~ T~ "5" 

z = 1 — y Plug into untouched equation 

-13y + 8(1 -j/) = - 55 Distribute 

— 13y + 8 — 8y = — 55 Combine like terms — 13?/ — 8y 

-21y + 8 = -55 Subtract8 



-21j/ = -63 Divide by -21 
^2T ^2T 

2/ = 3 We have our y\ Plug this into z = equations 

z = 1 — (3) Evaluate 

z = — 2 We have 2 , now find x from original equation. 

4x — 3(3) + 2( — 2) = — 29 Multiply and combine like terms 

4x-13 = -29 Add 13 
+ 13 +13 



4x = — 16 Divide by 4 
x = — 4 We have our x! 

(-4,3,-2) Our Solution! 



World View Note: Around 250, The Nine Chapters on the Mathematical Art 
were published in China. This book had 246 problems, and chapter 8 was about 
solving systems of equations. One problem had four equations with five variables! 

Just as with two variables and two equations, we can have special cases come up 
with three variables and three equations. The way we interpret the result is iden- 
tical. 



Example 184. 



5x-4y + 3z = -4: 
- lOrr + 8y — 6z = 8 We will eliminate x , start with first two equations 



154 



15x-12y + 9z = -12 

5x-4y + 3z = -4 LCM of 5 and - 10 is 10. 
-10x + 8y-6z = 8 

2(5x - 4y + Sz) = - 4(2) Multiply the first equation by 2 
10x-8y + 6z = -8 

10a; — 8y + 6z = — 8 Add this to the second equation, unchanged 
-10x + 8t/-6 = 8 

= A true statment 

Infinite Solutions Our Solution 



Example 185. 

3x-4y + z = 2 

— 9x + 12y — 3z = — 5 We will eliminates, starting with the first two equations 

4x — 2y — z = 3 

3x-4y + z = 2 The LCM of land -3 is 3 

- 9x + 12y - 3z = - 5 

3(3x — 4y + z) = (2)3 Multiply the first equation by 3 
9x-l2y + 3z = & 

9x — 12y + 3z = 6 Add this to the second equation, unchanged 
-9x + 12y-3z = -5 



= 1 A false statement 
No Solution Our Solution 



Equations with three (or more) variables are not any more difficult than two vari- 
ables if we are careful to keep our information organized and eliminate the same 
variable twice using two different pairs of equations. It is possible to solve each 
system several different ways. We can use different pairs of equations or eliminate 
variables in different orders, but as long as our information is organized and our 
algebra is correct, we will arrive at the same final solution. 



155 



4.4 Practice - Three Variables 



Solve each of the following systems of equation. 

I) a-2b + c = 5 2)2x + 3y = z-l 
2a + b-c = -l 3x = 8z-l 

3a + 3b-2c = -4 5y + 7z=-l 

3)3x + y-z = ll 4) x + y + z = 2 

x + 3y = z + 13 6x — Ay + 5z = 31 

x + y- 3z = ll hx + 2y + 2z = 13 

5) x + 6y + 3^ = 4 6) x — y + 2z = — 3 

2x + y + 2z = 3 x + 2y + 3z = 4 

3x-2y + z = 2x + y + z = -3 

7)x + y + z = 6 8)x + y — z = 

2x — y — z = — 3 x + 2y — 4z = 

x — 2y + 3z = 6 2x + y + z = 

9) x + y-z = 10)x + 2y-z = 4 

x — y — z = 4x — 3y + z = 8 

x + y + 2z = 5x-y = 12 

II) - 2x + y - 3z = 1 12) 4x + 12y + 16z = 4 
x — 4y + z = 6 3x + 4y + 5z = 3 
4x + lQy + 4z = 24 x + 8y+llz = l 

13) 2x + y - 3z = 14) 4x + I2y + 16z = 

x-4y + z = 3x + 4y + hz = 

4x + lQy + 4z = x + 8y+llz = Q 

15) 3a; + 2y + 2z = 3 16) p + g + r = 1 

re + 2?/ — 2 = 5 p + 2g + 3r = 4 

2x-4y + z = 4p + 5q + 6r = 7 

17) x-2y + 3z = 4 18) x + 2y-3z = 9 

2x-y + z = -l 2x- y + 2z = -8 

4x + y + z = l 3x — y — 4z = 3 

19) x-y + 2z = 20) 4x-7y + 3z = l 

x-2y + 3z = -l 3x + y-2z = 4 

2x — 2y + z = — 3 4x — 7y + 3z = 6 

21) 4x - 3y + 2z = 40 22) 3x + y - z = 10 

5x + 9y — 7z = 47 8x — y — 6z = — 3 

9x + 8y-3z = 97 5x-2y-5z = l 



156 



23) 3x + 3y - 2z = 13 
6x + 2y — hz = 13 
5a: — 2y — 5z = — 1 

25) 3x-4y + 2z = l 
2x + 3y — 3z = — 1 
x + l0y-8z = 7 

27) m + 6n + 3p = 8 
3m + 4n = — 3 
5m + 7n = l 

29) -2u> + 2o; + 2t/-22 = -10 
w+x+y+z=— 5 
3tf + 2x + 2?/ + 4^ = -ll 
to + 3x — 2y + 2z = — 6 

31) w + x + y + z = 2 
w + 2x + 2y + 4z = l 

— w + x — y — z = — 6 

— w + 3x + y — z = — 2 



24) 2x-3y + 5^ = l 
3x + 2y- z = 4 

4x + 7y — 7z = 7 

26) 2x + y = z 
4x + z = 4y 
y = x + l 

28) 3x + 2y = z + 2 
y = 1 — 2x 
3z = -2y 

30) -u> + 2;r-3?/ + ^ = -8 

— w + x + y — z = — 4 
^ + x + t/ + ^ = 22 

— w + x — y — z = — 14 

32) w + x — y + z = 

— w + 2x + 2y + z = 5 

— w + 3x + y — z = — 4 

— 2to + x + y — 3z = — 7 



157 



4.5 



Systems of Equations - Value Problems 



Objective: Solve value problems by setting up a system of equations. 

One application of system of equations are known as value problems. Value prob- 
lems are ones where each variable has a value attached to it. For example, if our 
variable is the number of nickles in a person's pocket, those nickles would have a 
value of five cents each. We will use a table to help us set up and solve value 
problems. The basic structure of the table is shown below. 





Number 


Value 


Total 


Item 1 








Item 2 








Total 









The first column in the table is used for the number of things we have. Quite 
often, this will be our variables. The second column is used for the that value 
each item has. The third column is used for the total value which we calculate by 
multiplying the number by the value. For example, if we have 7 dimes, each with 
a value of 10 cents, the total value is 7 • 10 = 70 cents. The last row of the table is 
for totals. We only will use the third row (also marked total) for the totals that 



158 



are given to use. This means sometimes this row may have some blanks in it. 
Once the table is filled in we can easily make equations by adding each column, 
setting it equal to the total at the bottom of the column. This is shown in the fol- 
lowing example. 



Example 186. 

In a child's bank are 11 coins that have a value of $1.85. The coins are either 
quarters or dimes. How many coins each does child have? 





Number 


Value 


Total 


Quarter 


q 


25 




Dime 


d 


10 




Total 












Number 


Value 


Total 


Quarter 


Q 


25 


25q 


Dime 


d 


10 


lOd 


Total 












Number 


Value 


Total 


Quarter 


Q 


25 


25q 


Dime 


d 


10 


lOd 


Total 


11 




185 



Using value table, use q for quarters, d for dimes 
Each quarter's value is 25 cents, dime's is 10 cents 



Multiply number by value to get totals 



We have 1 1 coins total. This is the number total. 
We have 1.85 for the final total, 
Write final total in cents (185) 
Because 25 and 10 are cents 



q + d = ll 
25q + 10d=185 



First and last columns are our equations by adding 
Solve by either addition or substitution. 



10(g + d) = (ll)(-10) 
-10g-10d=-110 



Using addition, multiply first equation by — 10 



10g-10d=-110 

25g + 10rf=185 

15q =75 

T5~ T5~ 

q = 5 



Add together equations 



Divide both sides by 15 



We have our q , number of quarters is 5 



(5) + d=ll 

-5 -5 

d = 6 



Plug into one of original equations 

Subtract 5 from both sides 

We have our d, number of dimes is 6 



159 



5 quarters and 6 dimes Our Solution 

World View Note: American coins are the only coins that do not state the 
value of the coin. On the back of the dime it says "one dime" (not 10 cents). On 
the back of the quarter it says "one quarter" (not 25 cents). On the penny it 
says "one cent" (not 1 cent). The rest of the world (Euros, Yen, Pesos, etc) all 
write the value as a number so people who don't speak the language can easily 
use the coins. 

Ticket sales also have a value. Often different types of tickets sell for different 
prices (values). These problems can be solve in much the same way. 

Example 187. 

There were 41 tickets sold for an event. Tickets for children cost $1.50 and tickets 
for adults cost $2.00. Total receipts for the event were $73.50. How many of each 
type of ticket were sold? 





Number 


Value 


Total 


Child 


c 


1.5 




Adult 


a 


2 




Total 












Number 


Value 


Total 


Child 


c 


1.5 


1.5c 


Adult 


a 


2 


2a 


Total 












Number 


Value 


Total 


Child 


c 


1.5 


1.5c 


Adult 


a 


2 


2a 


Total 


41 




73.5 



Using our value table, c for child, a for adult 
Child tickets have value 1.50, adult value is 2.00 
(we can drop the zeros after the decimal point) 



Multiply number by value to get totals 



We have 41 tickets sold. This is our number total 

The final total was 73.50 

Write in dollars as 1 .5 and 2 are also dollars 



c + a = 41 
1.5c + 2a = 73.5 



First and last columns are our equations by adding 
We can solve by either addition or substitution 



c + a = 41 We will solve by substitution. 
- c — c Solve for a by subtracting c 

Substitute into untouched equation 

Distribute 

Combine like terms 

Subtract 82 from both sides 

0.5c = — 8.5 Divide both sides by — 0.5 





a = 


41 -c 


1.5c + 2(41 


-c)- 


= 73.5 


1.5c + 82- 


-2c-- 


= 73.5 


-0.5c + 82: 


= 73.5 




-82 


-82 



160 



^03 ^03 

c=17 

o = 41-(17) 

a = 24 
17 child tickets and 24 adult tickets 



We have c, number of child tickets is 17 
Plug into a = equation to find a 
We have our a , number of adult tickets is 24 
Our Solution 



Some problems will not give us the total number of items we have. Instead they 
will give a relationship between the items. Here we will have statements such 
as "There are twice as many dimes as nickles". While it is clear that we need to 
multiply one variable by 2, it may not be clear which variable gets multiplied by 
2. Generally the equations are backwards from the English sentence. If there are 
twice as many dimes, than we multiply the other variable (nickels) by two. So the 
equation would be d = In. This type of problem is in the next example. 

Example 188. 

A man has a collection of stamps made up of 5 cent stamps and 8 cent stamps. 
There are three times as many 8 cent stamps as 5 cent stamps. The total value of 
all the stamps is $3.48. How many of each stamp does he have? 



Use value table, / for five cent stamp, and e for eight 
Also list value of each stamp under value column 





Number 


Value 


Total 


Five 


/ 


5 


5/ 


Eight 


3/ 


8 


24/ 


Total 






348 






Number 


Value 


Total 


Five 


/ 


5 


5/ 


Eight 


e 


8 


8e 


Total 












Number 


Value 


Total 


Five 


/ 


5 


5/ 


Eight 


e 


8 


8e 


Total 






348 



Multiply number by value to get total 



The final total was 338 (written in cents) 

We do not know the total number, this is left blank. 



e = 3/ 3 times as many 8 cent stamples as 5 cent stamps 

5 / + 8e = 348 Total column gives second equation 



5/ + 8(3/) =348 

5/ + 24/ = 348 

29/ = 348 

29" 29" 

/ = 12 

e = 3(12) 



Substitution, substitute first equation in second 

Multiply first 

Combine like terms 

Divide both sides by 39 

We have /. There are 12 five cent stamps 

Plug into first equation 



161 



e = 36 

12 five cent, 36 eight cent stamps 



We have e , There are 36 eight cent stamps 
Our Solution 



The same process for solving value problems can be applied to solving interest 
problems. Our table titles will be adjusted slightly as we do so. 





Invest 


Rate 


Interest 


Account 1 








Account 2 








Total 









Our first column is for the amount invested in each account. The second column 
is the interest rate earned (written as a decimal - move decimal point twice left), 
and the last column is for the amount of interset earned. Just as before, we mul- 
tiply the investment amount by the rate to find the final column, the interest 
earned. This is shown in the following example. 



Example 189. 

A woman invests $4000 in two accounts, one at 6% interest, the other at 9% 
interest for one year. At the end of the year she had earned $270 in interest. How 
much did she have invested in each account? 



Use our investment table, x and y for accounts 
Fill in interest rates as decimals 





Invest 


Rate 


Interest 


Account 1 


X 


0.06 




Account 2 


y 


0.09 




Total 












Invest 


Rate 


Interest 


Account 1 


X 


0.06 


0.06x 


Account 2 


y 


0.09 


0.09?/ 


Total 












Invest 


Rate 


Interest 


Account 1 


X 


0.06 


0.06x 


Account 2 


y 


0.09 


0.09?/ 


Total 


4000 




270 



Multiply across to find interest earned. 



Total investment is 4000, 
Total interest was 276 





x + y = 
0.06a; + 0.09?/ 


= 4000 
= 270 


0.06(: 


t + y) = 
-0.06z 


(4000) (- 
-0.06?/ = 


0.06) 
-240 



First and last column give our two equations 
Solve by either substitution or addition 

Use Addition, multiply first equation by — 0.06 



162 



0.06a; -0.06?/ = -240 
0.06x + 0.09^ = 270 



0.03 0.03 

y =1000 

x + 1000 = 4000 

-1000-1000 

£ = 3000 

$1000 at 9% and $3000 at 6% 



Add equations together 



0.03jy = 30 Divide both sides by 0.03 



We have y , $1000 invested at 9% 
Plug into original equation 
Subtract 1000 from both sides 
We have x , $3000 invested at 6% 
Our Solution 



The same process can be used to find an unknown interest rate. 

Example 190. 

John invests $5000 in one account and $8000 in an account paying 4% more in 
interest. He earned $1230 in interest after one year. At what rates did he invest? 



Our investment table. Use x for first rate 
The second rate is 4% higher, ori + 0.04 
Be sure to write this rate as a decimal! 





Invest 


Rate 


Interest 


Account 1 


5000 


X 




Account 2 


8000 


£ + 0.04 




Total 











Invest 


Rate 


Interest 


Account 1 


5000 


X 


5000a; 


Account 2 


8000 


x + 0.04 


8000^ + 320 


Total 











Invest 


Rate 


Interest 


Account 2 


5000 


X 


5000x 


Account 2 


8000 


x + 0.04 


8000^ + 320 


Total 






1230 



Multiply to fill in interest column. 
Be sure to distribute 8000(x + 0.04) 



Total interest was 1230. 



5000a; + 8000a; + 320 = 1230 

13000a; + 320 = 1230 

-320 -320 

13000a; = 910 



13000 13000 

a; = 0.07 

(0.07) + 0.04 

0.11 

$5000 at 7% and $8000 at 11% 



Last column gives our equation 
Combine like terms 
Subtract 320 from both sides 
Divide both sides by 13000 

We have our x , 7% interest 
Second account is 4% higher 
The account with $8000 is at 11% 
Our Solution 



163 



4.5 Practice - Value Problems 



Solve. 

1) A collection of dimes and quaters is worth $15.25. There are 103 coins in all. 
How many of each is there? 

2) A collection of half dollars and nickels is worth $13.40. There are 34 coins in 
all. How many are there? 

3) The attendance at a school concert was 578. Admission was $2.00 for adults 
and $1.50 for children. The total receipts were $985.00. How many adults and 
how many children attended? 

4) A purse contains $3.90 made up of dimes and quarters. If there are 21 coins in 
all, how many dimes and how many quarters were there? 

5) A boy has $2.25 in nickels and dimes. If there are twice as many dimes as 
nickels, how many of each kind has he? 

6) $3.75 is made up of quarters and half dollars. If the number of quarters 
exceeds the number of half dollars by 3, how many coins of each denomination 
are there? 

7) A collection of 27 coins consisting of nickels and dimes amounts to $2.25. How 
many coins of each kind are there? 

8) $3.25 in dimes and nickels, were distributed among 45 boys. If each received 
one coin, how many received dimes and how many received nickels? 

9) There were 429 people at a play. Admission was $1 each for adults and 75 
cents each for children. The receipts were $372.50. How many children and 
how many adults attended? 

10) There were 200 tickets sold for a women's basketball game. Tickets for 
students were 50 cents each and for adults 75 cents each. The total amount of 

money collected was $132.50. How many of each type of ticket was sold? 

11) There were 203 tickets sold for a volleyball game. For activity-card holders, 

the price was $1.25 each and for noncard holders the price was $2 each. The 
total amount of money collected was $310. How many of each type of ticket 
was sold? 

12) At a local ball game the hotdogs sold for $2.50 each and the hamburgers sold 
for $2.75 each. There were 131 total sandwiches sold for a total value of $342. 

How many of each sandwich was sold? 

13) At a recent Vikings game $445 in admission tickets was taken in. The cost of 
a student ticket was $1.50 and the cost of a non-student ticket was $2.50. A 
total of 232 tickets were sold. How many students and how many non- 
students attented the game? 

164 



14 
15 
16 
17 

18 

19 
20 
21 

22 

23 

24 



25 
26 

27 
28 
29 



A bank contains 27 coins in dimes and quarters. The coins have a total value 
of $4.95. Find the number of dimes and quarters in the bank. 

A coin purse contains 18 coins in nickels and dimes. The coins have a total 
value of $1.15. Find the number of nickels and dimes in the coin purse. 

A business executive bought 40 stamps for $9.60. The purchase included 25c 
stamps and 20c stamps. How many of each type of stamp were bought? 

A postal clerk sold some 15c stamps and some 25c stamps. Altogether, 15 
stamps were sold for a total cost of $3.15. How many of each type of stamps 
were sold? 

A drawer contains 15c stamps and 18c stamps. The number of 15c stamps is 
four less than three times the number of 18c stamps. The total value of all 
the stamps is $1.29. How many 15c stamps are in the drawer? 

The total value of dimes and quarters in a bank is $6.05. There are six more 
quarters than dimes. Find the number of each type of coin in the bank. 

A child's piggy bank contains 44 coins in quarters and dimes. The coins have 
a total value of $8.60. Find the number of quaters in the bank. 

A coin bank contains nickels and dimes. The number of dimes is 10 less than 
twice the number of nickels. The total value of all the coins is $2.75. Find the 
number of each type of coin in the bank. 

A total of 26 bills are in a cash box. Some of the bills are one dollar bills, and 
the rest are five dollar bills. The total amount of cash in the box is $50. Find 
the number of each type of bill in the cash box. 

A bank teller cashed a check for $200 using twenty dollar bills and ten dollar 
bills. In all, twelve bills were handed to the customer. Find the number of 
twenty dollar bills and the number of ten dollar bills. 

A collection of stamps consists of 22c stamps and 40c stamps. The number of 
22c stamps is three more than four times the number of 40c stamps. The 
total value of the stamps is $8.34. Find the number of 22c stamps in the 
collection. 

A total of $27000 is invested, part of it at 12% and the rest at 13%. The 
total interest after one year is $3385. How much was invested at each rate? 

A total of $50000 is invested, part of it at 5% and the rest at 7.5%. The total 
interest after one year is $3250. How much was invested at each rate? 

A total of $9000 is invested, part of it at 10% and the rest at 12%. The total 
interest after one year is $1030. How much was invested at each rate? 

A total of $18000 is invested, part of it at 6% and the rest at 9%. The total 
interest after one year is $1248. How much was invested at each rate? 

An inheritance of $10000 is invested in 2 ways, part at 9.5% and the 
remainder at 11%. The combined annual interest was $1038.50. How much 
was invested at each rate? 



165 



30) Kerry earned a total of $900 last year on his investments. If $7000 was 
invested at a certain rate of return and $9000 was invested in a fund with a 
rate that was 2% higher, find the two rates of interest. 

31) Jason earned $256 interest last year on his investments. If $1600 was invested 
at a certain rate of return and $2400 was invested in a fund with a rate that 
was double the rate of the first fund, find the two rates of interest. 

32) Millicent earned $435 last year in interest. If $3000 was invested at a certain 
rate of return and $4500 was invested in a fund with a rate that was 2% 
lower, find the two rates of interest. 

33) A total of $8500 is invested, part of it at 6% and the rest at 3.5%. The total 
interest after one year is $385. How much was invested at each rate? 

34) A total of $12000 was invested, part of it at 9% and the rest at 7.5%. The 
total interest after one year is $1005. How much was invested at each rate? 

35) A total of $15000 is invested, part of it at 8% and the rest at 11%. The total 
interest after one year is $1455. How much was invested at each rate? 

36) A total of $17500 is invested, part of it at 7.25% and the rest at 6.5%. The 
total interest after one year is $1227.50. How much was invested at each rate? 

37) A total of $6000 is invested, part of it at 4.25% and the rest at 5.75%. The 
total interest after one year is $300. How much was invested at each rate? 

38) A total of $14000 is invested, part of it at 5.5% and the rest at 9%. The total 
interest after one year is $910. How much was invested at each rate? 

39) A total of $11000 is invested, part of it at 6.8% and the rest at 8.2%. The 
total interest after one year is $797. How much was invested at each rate? 

40) An investment portfolio earned $2010 in interest last year. If $3000 was 
invested at a certain rate of return and $24000 was invested in a fund with a 
rate that was 4% lower, find the two rates of interest. 

41) Samantha earned $1480 in interest last year on her investments. If $5000 was 
invested at a certain rate of return and $11000 was invested in a fund with a 
rate that was two-thirds the rate of the first fund, find the two rates of 
interest. 

42) A man has $5.10 in nickels, dimes, and quarters. There are twice as many 
nickels as dimes and 3 more dimes than quarters. How many coins of each 
kind were there? 

43) 30 coins having a value of $3.30 consists of nickels, dimes and quarters. If 
there are twice as many quarters as dimes, how many coins of each kind were 
there? 

44) A bag contains nickels, dimes and quarters having a value of $3.75. If there 
are 40 coins in all and 3 times as many dimes as quarters, how many coins of 
each kind were there? 



166 



4.6 



Systems of Equations - Mixture Problems 



Objective: Solve mixture problems by setting up a system of equations. 

One application of systems of equations are mixture problems. Mixture problems 
are ones where two different solutions are mixed together resulting in a new final 
solution. We will use the following table to help us solve mixture problems: 





Amount 


Part 


Total 


Item 1 








Item 2 








Final 









The first column is for the amount of each item we have. The second column is 
labeled "part". If we mix percentages we will put the rate (written as a decimal) 
in this column. If we mix prices we will put prices in this column. Then we can 
multiply the amount by the part to find the total. Then we can get an equation 
by adding the amount and/or total columns that will help us solve the problem 
and answer the questions. 

These problems can have either one or two variables. We will start with one vari- 
able problems. 



Example 191. 

A chemist has 70 mL of a 50% methane solution. How much of a 80% solution 
must she add so the final solution is 60% methane? 





Amount 


Part 


Total 


Start 


70 


0.5 




Add 


X 


0.8 




Final 









Set up the mixture table. We start with 70, but 
don't know how much we add, that is x. The part 
is the percentages, 0.5 for start, 0.8 for add. 



167 





Amount 


Part 


Total 


Start 


70 


0.5 




Add 


X 


0.8 




Final 


70 + x 


0.6 





Add amount column to get final amount . The 
part for this amount is 0.6 because we want the 
final solution to be 60% methane. 





Amount 


Part 


Total 


Start 


70 


0.5 


35 


Add 


X 


0.8 


0.8x 


Final 


70 + x 


0.6 


42 + 0.6a; 



Multiply amount by part to get total. 

be sure to distribute on the last row: (70 + a;)0.6 



35 + 0.8x = 42 + 0.6x 

— 0.6a; — 0.6a; 

35 + 0.2x = 42 

- 35 - 35 

0.2a; = 7 

O 0T2 

x = 35 

35 mL must be added 



The last column is our equation by adding 
Move variables to one side, subtract 0.6a: 
Subtract 35 from both sides 

Divide both sides by 0.2 

We have our x ! 
Our Solution 



The same process can be used if the starting and final amount have a price 
attached to them, rather than a percentage. 



Example 192. 

A coffee mix is to be made that sells for $2.50 by mixing two types of coffee. The 
cafe has 40 mL of coffee that costs $3.00. How much of another coffee that costs 
$1.50 should the cafe mix with the first? 





Amount 


Part 


Total 


Start 


40 


3 




Add 


X 


1.5 




Final 









Set up mixture table. We know the starting 
amount and its cost, $3. The added amount 
we do not know but we do know its cost is $1.50. 





Amount 


Part 


Total 


Start 


40 


3 




Add 


X 


1.5 




Final 


40 + x 


2.5 





Add the amounts to get the final amount. 
We want this final amount to sell for $2.50. 



168 





Amount 


Part 


Total 


Start 


40 


3 


120 


Add 


X 


1.5 


1.5a: 


Final 


40 + x 


2.5 


100 + 2.5x 



Multiply amount by part to get the total. 

Be sure to distribute on the last row (40 + x)2.5 



120 



1.5a:- 
1.5a; 



100- 



2.5a; 

1.5a: 



120: 

100 



100 
100 



x 



20 = a: 
20mL must be added. 



Adding down the total column gives our equation 
Move variables to one side by subtracting 1.5a; 
Subtract 100 from both sides 

We have our x. 
Our Solution 



World View Note: Brazil is the world's largest coffee producer, producing 2.59 
million metric tons of coffee a year! That is over three times as much coffee as 
second place Vietnam! 

The above problems illustrate how we can put the mixture table together and get 
an equation to solve. However, here we are interested in systems of equations, 
with two unknown values. The following example is one such problem. 



Example 193. 

A farmer has two types of milk, one that is 24% butterfat and another which is 
18% butterfat. How much of each should he use to end up with 42 gallons of 20% 
butterfat? 





Amount 


Part 


Total 


Milkl 


X 


0.24 




Milk 2 


y 


0.18 




Final 


42 


0.2 








Amount 


Part 


Total 


Milkl 


X 


0.24 


0.24a: 


Milk 2 


y 


0.18 


0.18?/ 


Final 


42 


0.2 


8.4 



We don't know either start value, but we do know 
final is 42. Also fill in part column with percentage 
of each type of milk including the final solution 



Multiply amount by part to get totals. 



x + y = 42 
0.24a: + 0.18?/ = 8.4 



The amount column gives one equation 
The total column gives a second equation. 



169 



0.18(a; + y) = (42)(-0.18) 
- 0.18a; — 0.182/= -7.56 



Use addition. Multiply first equation by — 0.18 



0.18a; — 0.182/= — 7.56 
0.24a; + 0.18t/ = 8.4 



0.06a; = 


0.84 


0.06 


0.06 


X 


= 14 


(14) + ?/ 


= 42 


-14 


-14 



2/ = 28 
14 gal of 24% and 28 gal of 18% 



Add the equations together 

Divide both sides by 0.06 

We have our x , 14 gal of 24% butterfat 
Plug into original equation to find y 
Subtract 14 from both sides 
We have our y , 28 gal of 18% butterfat 
Our Solution 



The same process can be used to solve mixtures of prices with two unknowns. 



Example 194. 

In a candy shop, chocolate which sells for $4 a pound is mixed with nuts which 
are sold for $2.50 a pound are mixed to form a chocolate-nut candy which sells 
for $3.50 a pound. How much of each are used to make 30 pounds of the mix- 
ture? 





Amount 


Part 


Total 


Chocolate 


c 


4 




Nut 


n 


2.5 




Final 


30 


3.5 








Amount 


Part 


Total 


Chocolate 


c 


4 


4c 


Nut 


n 


2.5 


2.5n 


Final 


30 


3.5 


105 



Using our mixture table, use c and n for variables 
We do know the final amount (30) and price, 
include this in the table 



Multiply amount by part to get totals 



c + n = 30 
4c + 2.5n=105 



First equation comes from the first column 
Second equation comes from the total column 



c + n = 30 

— n — n 

c = 30 — n 



We will solve this problem with substitution 
Solve for cby subtracting n from the first equation 



170 



4(30 -n) + 2.5n = 105 

120-4n + 2.5n = 105 

120-1.5n=105 

- 120 - 120 



1.5n : 



15 



-1.5 -1.5 
n = 10 

c = 30-(10) 

c = 20 

10 lbs of nuts and 20 lbs of chocolate 



Substitute into untouched equation 
Distribute 
Combine like terms 
Subtract 120 from both sides 
Divide both sides by — 1 .5 

We have our n , 1 lbs of nuts 
Plug into c = equation to find c 
We have our c, 20 lbs of chocolate 
Our Solution 



With mixture problems we often are mixing with a pure solution or using water 
which contains none of our chemical we are interested in. For pure solutions, the 
percentage is 100% (or 1 in the table). For water, the percentage is 0%. This is 
shown in the following example. 



Example 195. 

A solution of pure antifreeze is mixed with water to make a 65% antifreeze solu- 
tion. How much of each should be used to make 70 L? 





Amount 


Part 


Final 


Antifreeze 


a 


1 




Water 


w 







Final 


70 


0.65 








Amount 


Part 


Final 


Antifreeze 


a 


1 


a 


Water 


w 








Final 


70 


0.65 


45.5 



We use a and w for our variables. Antifreeze 
is pure, 100% or 1 in our table, written as a 
decimal. Water has no antifreeze, its 
percentage is 0. We also fill in the final percent 



Multiply to find final amounts 



o + w = 70 

a = 45.5 

(45.5)+w; = 70 

-45.5 -45.5 

w; = 24.5 

45. 5 L of antifreeze and 24. 5 L of water 



First equation comes from first column 
Second equation comes from second column 
We have a , plug into to other equation 
Subtract 45.5 from both sides 
We have our w 
Our Solution 



171 



4.6 Practice - Mixture Problems 



Solve. 



1) A tank contains 8000 liters of a solution that is 40% acid. How much water 
should be added to make a solution that is 30% acid? 

2) How much antifreeze should be added to 5 quarts of a 30% mixture of 
antifreeze to make a solution that is 50% antifreeze? 

3) Of 12 pounds of salt water 10% is salt; of another mixture 3% is salt. How 
many pounds of the second should be added to the first in order to get a 
mixture of 5% salt? 

4) How much alcohol must be added to 24 gallons of a 14% solution of alcohol in 
order to produce a 20% solution? 

5) How many pounds of a 4% solution of borax must be added to 24 pounds of a 
12% solution of borax to obtain a 10% solution of borax? 

6) How many grams of pure acid must be added to 40 grams of a 20% acid 

solution to make a solution which is 36% acid? 

7) A 100 LB bag of animal feed is 40% oats. How many pounds of oats must be 
added to this feed to produce a mixture which is 50% oats? 

8) A 20 oz alloy of platinum that costs $220 per ounce is mixed with an alloy 
that costs $400 per ounce. How many ounces of the $400 alloy should be used 
to make an alloy that costs $300 per ounce? 

9) How many pounds of tea that cost $4.20 per pound must be mixed with 12 lb 
of tea that cost $2.25 per pound to make a mixture that costs $3.40 per 
pound? 

10) How many liters of a solvent that costs $80 per liter must be mixed with 6 L 
of a solvent that costs $25 per liter to make a solvent that costs $36 per liter? 

11) How many kilograms of hard candy that cost $7.50 per kilogram must be 

mixed with 24 kg of jelly beans that cost $3.25 per kilogram to make a 
mixture that sells for $4.50 per kilogram? 

12) How many kilograms of soil supplement that costs $7.00 per kilogram must 
be mixed with 20 kg of aluminum nitrate that costs $3.50 per kilogram to 
make a fertilizer that costs $4.50 per kilogram? 

13) How many pounds of lima beans that cost 90c per pound must be mixed with 
16 lb of corn that cost 50c per pound to make a mixture of vegetables that 

costs 65c per pound? 

14) How many liters of a blue dye that costs $1.60 per liter must be mixed with 
18 L of anil that costs $2.50 per liter to make a mixture that costs $1.90 per 
liter? 

15) Solution A is 50% acid and solution B is 80% acid. How much of each should 
be used to make lOOcc. of a solution that is 68% acid? 



172 



16) A certain grade of milk contains 10% butter fat and a certain grade of cream 
60% butter fat. How many quarts of each must be taken so as to obtain a 
mixture of 100 quarts that will be 45% butter fat? 

17) A farmer has some cream which is 21% butterfat and some which is 15% 
butter fat. How many gallons of each must be mixed to produce 60 gallons of 
cream which is 19% butterfat? 

18) A syrup manufacturer has some pure maple syrup and some which is 85% 
maple syrup. How many liters of each should be mixed to make 150L which 
is 96% maple syrup? 

19) A chemist wants to make 50ml of a 16% acid solution by mixing a 13% acid 
solution and an 18% acid solution. How many milliliters of each solution 
should the chemist use? 

20) A hair dye is made by blending 7% hydrogen peroxide solution and a 4% 
hydrogen peroxide solution. How many mililiters of each are used to make a 

300 ml solution that is 5% hydrogen peroxide? 

21) A paint that contains 21% green dye is mixed with a paint that contains 15% 
green dye. How many gallons of each must be used to make 60 gal of paint 
that is 19% green dye? 

22) A candy mix sells for $2.20 per kilogram. It contains chocolates worth $1.80 
per kilogram and other candy worth $3.00 per kilogram. How much of each 
are in 15 kilograms of the mixture? 

23) To make a weed and feed mixture, the Green Thumb Garden Shop mixes 

fertilizer worth $4.00/lb. with a weed killer worth $8.00/lb. The mixture 
will cost $6.00/lb. How much of each should be used to prepare 500 lb. of 
the mixture? 

24) A grocer is mixing 40 cent per lb. coffee with 60 cent per lb. coffee to make a 
mixture worth 54c per lb. How much of each kind of coffee should be used to 
make 70 lb. of the mixture? 

25) A grocer wishes to mix sugar at 9 cents per pound with sugar at 6 cents per 
pound to make 60 pounds at 7 cents per pound. What quantity of each must 
he take? 

26) A high-protein diet supplement that costs $6.75 per pound is mixed with a 
vitamin supplement that costs $3.25 per pound. How many pounds of each 
should be used to make 5 lb of a mixture that costs $4.65 per pound? 

27) A goldsmith combined an alloy that costs $4.30 per ounce with an alloy that 
costs $1.80 per ounce. How many ounces of each were used to make a mixture 
of 200 oz costing $2.50 per ounce? 

28) A grocery store offers a cheese and fruit sampler that combines cheddar cheese 
that costs $8 per kilogram with kiwis that cost $3 per kilogram. How many 
kilograms of each were used to make a 5 kg mixture that costs $4.50 per 
kilogram? 

173 



29) The manager of a garden shop mixes grass seed that is 60% rye grass with 70 

lb of grass seed that is 80% rye grass to make a mixture that is 74% rye 
grass. How much of the 60% mixture is used? 

30) How many ounces of water evaporated from 50 oz of a 12% salt solution to 

produce a 15% salt solution? 

31) A caterer made an ice cream punch by combining fruit juice that cost $2.25 

per gallon with ice cream that costs $3.25 per gallon. How many gallons of 
each were used to make 100 gal of punch costing $2.50 per pound? 

32) A clothing manufacturer has some pure silk thread and some thread that is 

85% silk. How many kilograms of each must be woven together to make 75 
kg of cloth that is 96% silk? 

33) A carpet manufacturer blends two fibers, one 20% wool and the second 50% 
wool. How many pounds of each fiber should be woven together to produce 
600 lb of a fabric that is 28% wool? 

34) How many pounds of coffee that is 40% Java beans must be mixed with 80 lb 
of coffee that is 30% Java beans to make a coffee blend that is 32% Java 
beans? 

35) The manager of a specialty food store combined almonds that cost $4.50 per 

pound with walnuts that cost $2.50 per pound. How many pounds of each 
were used to make a 100 lb mixture that cost $3.24 per pound? 

36) A tea that is 20% jasmine is blended with a tea that is 15% jasmine. How 

many pounds of each tea are used to make 5 lb of tea that is 18% jasmine? 

37) How many ounces of dried apricots must be added to 18 oz of a snack mix 

that contains 20% dried apricots to make a mixture that is 25% dried 
apricots? 

38) How many mililiters of pure chocolate must be added to 150 ml of chocolate 
topping that is 50% chocolate to make a topping that is 75% chocolate? 

39) How many ounces of pure bran flakes must be added to 50 oz of cereal that 

is 40% bran flakes to produce a mixture that is 50% bran flakes? 

40) A ground meat mixture is formed by combining meat that costs $2.20 per 
pound with meat that costs $4.20 per pound. How many pounds of each 
were used to make a 50 lb mixture tha costs $3.00 per pound? 

41) How many grams of pure water must be added to 50 g of pure acid to make a 

solution that is 40% acid? 

42) A lumber company combined oak wood chips that cost $3.10 per pound with 
pine wood chips that cost $2.50 per pound. How many pounds of each were 
used to make an 80 lb mixture costing $2.65 per pound? 

43) How many ounces of pure water must be added to 50 oz of a 15% saline 

solution to make a saline solution that is 10% salt? 



174 



175 



Chapter 5 : Polynomials 



5.1 Exponent Properties 177 

5.2 Negative Exponents 183 

5.3 Scientific Notation 188 

5.4 Introduction to Polynomials 192 

5.5 Multiply Polynomials 196 

5.6 Multiply Special Products 201 

5.7 Divide Polynomials 205 



176 



5.1 

Polynomials - Exponent Properties 

Objective: Simplify expressions using the properties of exponents. 

Problems with expoenents can often be simplified using a few basic exponent 
properties. Exponents represent repeated multiplication. We will use this fact to 
discover the important properties. 

World View Note: The word exponent comes from the Latin "expo" meaning 
out of and "ponere" meaning place. While there is some debate, it seems that the 
Babylonians living in Iraq were the first to do work with exponents (dating back 
to the 23rd century BC or earlier!) 

Example 196. 

a 3 a 2 Expand exponents to multiplication problem 
(a a a) (a a) Now we have 5 a's being multiplied together 
o 5 Our Solution 

A quicker method to arrive at our answer would have been to just add the expo- 
nents: a 3 a 2 = a 3+2 = a 5 This is known as the product rule of exponents 

Product Rule of Exponents: a m a n = a m+n 

The product rule of exponents can be used to simplify many problems. We will 
add the exponent on like variables. This is shown in the following examples 

Example 197. 

3 2 • 3 6 • 3 Same base, add the exponents 2 + 6 + 1 
3 9 Our Solution 

Example 198. 

2x 3 y 5 z ■ 5xy 2 z 3 Multiply 2 • 5, add exponents on x, yand z 
10x 4: y 7 z A Our Solution 

Rather than multiplying, we will now try to divide with exponents 

Example 199. 

a 5 



a 2 
aaaaa 



Expand exponents 

Divide out two of the a's 

aa 

aaa Convert to exponents 
a 3 Our Solution 

177 



A quicker method to arrive at the solution would have been to just subtract the 
exponents, ^ = a 5 ~ 2 = a 3 . This is known as the quotient rule of exponents. 



a m 



Quotient Rule of Exponents: = a m n 

a n 

The quotient rule of exponents can similarly be used to simplify exponent prob- 
lems by subtracting exponents on like variables. This is shown in the following 
examples. 



Example 200. 



? 13 

— g- Same base, subtract the exponents 
7 8 Our Solution 



Example 201. 



5a 3 b"c 



5^2 

Subtract exponents on a, b and c 



2ab 3 c 

5 

—a 2 b 2 c Our Solution 

A third property we will look at will have an exponent problem raised to a second 
exponent. This is investigated in the following example. 

Example 202. 

( a 2 ) This means we have a 2 three times 
a 2 ■ a 2 ■ a 2 Add exponents 
a 6 Our solution 

A quicker method to arrive at the solution would have been to just multiply the 
exponents, (a 2 ) 3 = a 2 ' 3 = a 6 . This is known as the power of a power rule of expo- 
nents. 

Power of a Power Rule of Exponents: (a m ) n = a mn 

This property is often combined with two other properties which we will investi- 
gate now. 

Example 203. 

(ab) 3 This means we have (a b) three times 
(ab)(ab)(ab) Three a's and three b's can be written with exponents 
a 3 b 3 Our Solution 



178 



A quicker method to arrive at the solution would have been to take the exponent 
of three and put it on each factor in parenthesis, (ab) 3 = a 3 b 3 . This is known as 
the power of a product rule or exponents. 

Power of a Product Rule of Exponents: (ab) 7 " = a m b m 

It is important to be careful to only use the power of a product rule with multipli- 
cation inside parenthesis. This property does NOT work if there is addition or 
subtraction. 



Warning 204. 

(a + b) m ^ a m + b m These are NOT equal, beware of this error! 



Another property that is very similar to the power of a product rule is considered 
next. 



Example 205. 



b 



a\/a\/a 
~b)\~b)\~b 



b 3 



This means we have the fraction three timse 



Multiply fractions across the top and bottom, using exponents 



Our Solution 



A quicker method to arrive at the solution would have been to put the exponent 

Q 3 

on every factor in both the numerator and denominator, (^) = |g-. This is known 
as the power of a quotient rule of exponents. 



Power of a Quotient Rule of Exponents: ( — 



The power of a power, product and quotient rules are often used together to sim- 
plify expressions. This is shown in the following examples. 

Example 206. 

(x 3 yz 2 ) 4 Put the exponent of 4 on each factor, multiplying powers 
x 12 y 4 z 8 Our solution 



179 



Example 207. 



/ a 3 b V 

I -gjg ) Put the exponent of 2 on each factor, multiplying powers 



67,2 



c s d 10 



Our Solution 



As we multiply exponents its important to remember these properties apply to 
exponents, not bases. An expressions such as 5 3 does not mean we multipy 5 by 3, 
rather we multiply 5 three times, 5x5x5 = 125. This is shown in the next 
example. 

Example 208. 

(4x 2 y 5 ) 3 Put the exponent of 3 on each factor, multiplying powers 
A 3 x 6 y 15 Evaluate 4 3 
64x 6 y 15 Our Solution 

In the previous example we did not put the 3 on the 4 and multipy to get 12, this 
would have been incorrect. Never multipy a base by the exponent. These proper- 
ties pertain to exponents only, not bases. 

In this lesson we have discussed 5 different exponent properties. These rules are 
summarized in the following table. 

Rules of Exponents 



Product Rule of Exponents 


a m a n = a m+n 


Quotient Rule of Exponents 


m 

_ „m — n 


Power of a Power Rule of Exponents 


(a m ) n = a mn 


Power of a Product Rule of Exponents 


(ab) m = a m b m 


Power of a Quotient Rule of Exponents 


/a\ m a m 
\b) ~b^ 



These five properties are often mixed up in the same problem. Often there is a bit 
of flexibility as to which property is used first. However, order of operations still 
applies to a problem. For this reason it is the suggestion of the auther to simplify 
inside any parenthesis first, then simplify any exponents (using power rules), and 
finally simplify any multiplication or division (using product and quotient rules). 
This is illustrated in the next few examples. 



Example 209. 

(4x 3 y ■ 5x 4 y 2 ) 3 
(2CteV) 3 

2ovy 

800Cte 2 y 



In parenthesis simplify using product rule, adding exponents 
With power rules, put three on each factor, multiplying exponents 
Evaluate 20 3 
Our Solution 



180 



Example 210. 



7a 3 (2a 4 ) 3 Parenthesis are already simplified, next use power rules 
7a 3 (8a 12 ) Using product rule, add exponents and multiply numbers 
56a 15 Our Solution 



Example 


! 211. 


3a ; 


5 6-10a 4 6 3 




2a% 2 




30a 7 6 4 



Simplify numerator with product rule, adding exponents 
Now use the quotient rule to subtract exponents 



2a% 2 

15a 3 6 2 Our Solution 



Example 212. 



3m n TT . . . 

- — U se power rule in denominator 

(m z n A ) 6 

3m 8 n 12 



m 6 n 9 



Use quotient rule 



3mn Our solution 



Example 213. 



2 



( « 5h7 — ) Simplify inside parenthesis first, using power rule in numerator 

^r-= — - ) Simplify numerator using product rule 

6a b b' J 

24a 13 b 8 \ 2 
„ gl _ Simplify using the quotient rule 

ocro' / 

(4a 8 6) 2 Now that the parenthesis are simplified, use the power rules 
16a 16 fe 2 Our Solution 



Clearly these problems can quickly become quite involved. Remember to follow 
order of operations as a guide, simplify inside parenthesis first, then power rules, 
then product and quotient rules. 



181 



5.1 Practice - Exponent Properties 



Simplify. 

1) 4.44.44 

3) 4-2 2 

5) 3m • 4mn 

7) 2m 4 n 2 ■ 4nm 2 

9) (3 3 ) 4 



11 

13; 

17; 
19; 
21 
23; 

25; 
27; 

29; 
31; 

33) 
35) 
37) 
39) 
41) 
43) 



(4 4 ) 2 

(2uV) 2 

(2a 4 ) 4 

l! 
43 

& 

3 

3nm 2 
3n 

3xy 3 

(x 3 y 4 -2x 2 y 3 ) 2 
2x{x 4 y 4 ) 4 

2x 7 y 5 
3x 3 y ■ 4x 2 y 3 

(2x) 3 



2y 17 Y 3 

(2^^4)4 ) 

( 2m n 4 • 2m 4 n 4 A 

V ran 4 J 

2xy 5 -2x 2 y 3 
2xy 4 ■ y 3 

q 3 r 2 ■(2p 2 q 2 r 3 ) 2 
2p~ 3 

( zy 3 -z 3 x 4 y 4 \ 4 

V x 3 i/ 3 2 3 y 

2z 2 y 2 z 6 • 2,zirV 



2) 4-4 4 -4 2 
4) 3-3 3 -3 2 
6) 3x-4x 2 



2 4 2 

x y ■ xy 



10 
12 
14 
16 

18 
20 

22 

24 
26 

28 
30 
32 
34 
36 
38 
40 

42 



(4 3)4 
(3 2 ) 3 

(xy) 3 

(2xy) 4 

3^ 
33 

3^ 

3 



Axy 



xy 
Axy 

(u 2 v 2 • 2u 4 ) 3 

3vu 5 ■ 2v 3 
uv 2 ■ 2u 3 v 

26a 7 • 2o 4 



ba 2 • 3a 3 b 4 
2a 2 fe 2 a 7 



(ba*) 2 

yx 2 -{y 4 ) 2 

2y 4 

n 3 (n 4 ) 2 
2mn 

(2y 3 x 2 ) 2 
2x 2 y 4 ■ x 2 

2x 4 y 5 -2z 10 x 2 y 7 
(xy 2 z 2 ) 4 



2g 3 p 3 r 4 -2p 3 x4 



(grp 3 ) 2 



2-/.r V 2 



(x 2 z 3 ) 



182 



5.2 

Polynomials - Negative Exponents 

Objective: Simplify expressions with negative exponents using the 
properties of exponents. 

There are a few special exponent properties that deal with exponents that are not 
positive. The first is considered in the following example, which is worded out 2 
different ways: 

Example 214. 

a 3 

— ^ Use the quotient rule to subtract exponents 

a 6 

a° Our Solution, but now we consider the problem a the second way: 
Rewrite exponents as repeated multiplication 



a 3 



a 

aaa 
aaa 

1 



Reduce out all the a's 



1 Our Solution, when we combine the two solutions we get: 
a°=l Our final result . 

This final result is an imporant property known as the zero power rule of expo- 
nents 

Zero Power Rule of Exponents: a = 1 

Any number or expression raised to the zero power will always be 1. This is illus- 
trated in the following example. 



Example 215. 



(3x 2 ) Zero power rule 
1 Our Solution 



Another property we will consider here deals with negative exponents. Again we 
will solve the following example two ways. 

183 



a 3 



a : > 



a 3 



a" 

aaa 
aaaaa 

1 



Example 216. 

Using the quotient rule, subtract exponents 
r 

a~ 2 Our Solution, but we will also solve this problem another way. 
Rewrite exponents as repeated multiplication 
Reduce three a's out of top and bottom 

Simplify to exponents 

Our Solution, putting these solutions together gives: 
a 1 

a~ 2 = ^r Our Final Solution 
a 2 

This example illustrates an important property of exponents. Negative exponents 
yield the reciprocal of the base. Once we take the reciprical the exponent is now 
positive. Also, it is important to note a negative exponent does not mean the 
expression is negative, only that we need the reciprocal of the base. Following are 
the rules of negative exponents 

a- m = — 
m 



aa 
1 



m 



Rules of Negative Exponets: — - — = a 



a\~ m _ b" 1 
b) ~~a™ 

Negative exponents can be combined in several different ways. As a general rule if 
we think of our expression as a fraction, negative exponents in the numerator 
must be moved to the denominator, likewise, negative exponents in the denomi- 
nator need to be moved to the numerator. When the base with exponent moves, 
the exponent is now positive. This is illustrated in the following example. 

Example 217. 

o. b~ c 

Negative exponents on b, d, and e need to flip 



2d- l e-\f 2 

a 3 cde A 
2b 2 f 2 



Our Solution 



184 



As we simplified our fraction we took special care to move the bases that had a 
negative exponent, but the expression itself did not become negative because of 
those exponents. Also, it is important to remember that exponents only effect 
what they are attached to. The 2 in the denominator of the above example does 
not have an exponent on it, so it does not move with the d. 

We now have the following nine properties of exponents. It is important that we 
are very familiar with all of them. 

Properties of Exponents 



a m a n = a m+n 



a 



a' 



= a 



(a 


,b) 


m 


a m V 


( 


a 
b 


\ m 


_ a m 
~~b™ 



a' 



= a 



a 



(a m ) n = a t 



a° = l 



a' 



World View Note: Nicolas Chuquet, the French mathematician of the 15th cen- 
tury wrote 12 lm to indicate 12a; -1 . This was the first known use of the negative 
exponent. 

Simplifying with negative exponents is much the same as simplifying with positive 
exponents. It is the advice of the author to keep the negative exponents until the 
end of the problem and then move them around to their correct location (numer- 
ator or denominator). As we do this it is important to be very careful of rules for 
adding, subtracting, and multiplying with negatives. This is illustrated in the fol- 
lowing examples 



Example 218. 



4x b y 3 -Sx 3 y 2 
6x~ 5 y 3 

12x- 2 y- 5 

5,,3 



6x~ 5 y 



Simplify numerator with product rule, adding exponents 



Quotient rule to subtract exponets, be careful with negatives! 



(_ 2 )-(-5) = (-2) + 5 = 3 
(-5)-3 = (-5) + (-3) = - 



2x y Negative exponent needs to move down to denominator 



2x 3 



Our Solution 



185 



Example 219. 

(3ab 3 ) _2 ab -3 In numerator, use power rule with — 2, multiplying exponents 
2a _4 o° In denominator, b° = 1 

S- 2 a ~ 2 b~ 6 ab~ 3 

In numerator, use product rule to add exponents 



Use quotient rule to subtract exponents, be careful with negatives 
(-l)-(-4) = (-l)+4 = 3 

Move 3 and b to denominator because of negative exponents 
Evaluate 3 2 2 
Our Solution 



2a" 4 


Z^a-^b- 9 


2a" 4 


3" 2 a 3 6- 9 


2 


a 3 


3 2 26 9 


a 3 



186 9 



In the previous example it is important to point out that when we simplified 3 -2 
we moved the three to the denominator and the exponent became positive. We 
did not make the number negative! Negative exponents never make the bases neg- 
ative, they simply mean we have to take the reciprocal of the base. One final 
example with negative exponents is given here. 

Example 220. 

3x~ 2 y 5 z 3 ■ 6x~ 6 y~ 2 z~ 3 \ " In numerator, use product rule, adding exponents 



9(x 2 y~ 2 )~ 3 J In denominator, use power rule, multiplying exponets 

18x- 8 y 3 z°\~ 3 

— _ — J Use quotient rule to subtract exponents, 

ux y j 

be careful with negatives: 
(_ 8 )-(-6) = (-8) + 6 = -2 
3-6 = 3 + (-6) = -3 
(2x~ 2 y~ 3 z°)~ 3 Parenthesis are done, use power rule with — 3 
2~ 3 x 6 y 9 z° Move 2 with negative exponent down and z° = 1 



^-f- Evaluate 2 3 



x 6 y 9 

— — Our Solution 

8 



186 



5.2 Practice - Negative Exponents 

Simplify. Your answer should contain only positive expontents. 

1) 2x V 2 • (2xy 3 ) 4 2) 2a~ 2 b- 3 ■ (2a°6 4 ) 4 

3) (a 4 6" 3 ) 3 • 2a 3 6" 2 4) 2x 3 y 2 ■ (2x 3 ) 

5) (2xV) 4 £- 4 6) (m°n 3 -2m- 3 n- 3 ) 

7) (x 3 ?/ 4 ) 3 • x- V 8) 2m- ! n- 3 • (2m" 1 n- 3 ) 4 

9) 2 '" V 10) 3y3 

^/ Q„-3„,3 q„0 W / 



13 

15 
17 

19 
21 
23 
25 
27 
29 
31 
33 
35 
37 
39 



,,3 ^-3„,2 



3a;-V-3z ; 3yx 3 -2x 4 y- 3 



4xy- 3 -x- 4 y° 19^ 3a: 3 ;/ 2 



4 y -i " y 4y- 2 -3x" 2 y- 

2a;j/ 2 -4x 3 y- 4 



2«V-2u?; ' 4x~ 4 y~ 4 -4x 

" 2 -^ 2s -y 

4u°v 3 -3v 2 ' 4yx 2 

2y -.on (a 4 ) 4 

(*'V) 4 18 ) "^~ 



(^?) 4 20) (% 



\-2 



2nm 4 „„>. 2j/ 2 



(2m 2 n 2 ) 4 ' (^V)" 4 

(2mn) 4 .n 2x~ 3 



m°n 2 ' (x 4 y 3 ) 



24 )t^ 



-31-1 



(*V) 3 ZK> > (xyf)" 1 



2 M - 2 i) 3 -(2 M ^ 4 )- 1 oo ^ 2yx 2 -x" 2 



2U~ 4 V° ^°) fW),,4\-l 



C-^A 3 OM u- 3 v- 4 

^ y 4 ' 6U > 2t,(2u-M)° 

y(2sV) 2 6 _i 

2 ^° 32 ) (2a 4 6")"-2a" 3 6 2 
^^ 0/A 2b 4 c- 2 -(2& 3 c 2 )- 4 

2*V*- 2 -(*» a ) 4 34) a _ 2fe4 

2kh°-2h- 3 k° { (2x- 3 y°z- 1 ) 3 -x- 3 y 2 

(2kj 3 ) 2 J®) { ^3 

(cfe 3 ) 2 -2a" 3 6 2 „^ 2 g 4 -m 2 p 2 9 4 

(a 3 fc- 2 c 3 ) 3 ' (2m- 4 p 2 ) 3 

(yx-^ 2 )" 1 , n s 2mprt- 3 



40) 



2 y 3 z -i "J (m°n-V) 3 -2n 2 p° 



187 



5.3 

Polynomials - Scientific Notation 

Objective: Multiply and divide expressions using scientific notation and 
exponent properties. 

One application of exponent properties comes from scientific notation. Scientific 
notation is used to represent really large or really small numbers. An example of 
really large numbers would be the distance that light travels in a year in miles. 
An example of really small numbers would be the mass of a single hydrogen atom 
in grams. Doing basic operations such as multiplication and division with these 
numbers would normally be very combersome. However, our exponent properties 
make this process much simpler. 

First we will take a look at what scientific notation is. Scientific notation has two 
parts, a number between one and ten (it can be equal to one, but not ten), and 
that number multiplied by ten to some exponent. 

Scientific Notation: a X 10 b where 1 ^ a < 10 

The exponent, b, is very important to how we convert between scientific notation 
and normal numbers, or standard notation. The exponent tells us how many 
times we will multiply by 10. Multiplying by 10 in affect moves the decimal point 
one place. So the exponent will tell us how many times the exponent moves 
between scientific notation and standard notation. To decide which direction to 
move the decimal (left or right) we simply need to remember that positive expo- 
nents mean in standard notation we have a big number (bigger than ten) and neg- 
ative exponents mean in standard notation we have a small number (less than 
one). 

Keeping this in mind, we can easily make conversions between standard notation 
and scientific notation. 

Example 221. 

Convert 14, 200 to scientific notation Put decimal after first nonzero number 

1 .42 Exponent is how many times decimal moved, 4 

x 10 4 Positive exponent, standard notation is big 

1.42 x 10 4 Our Solution 

Example 222. 

Convert 0.0042 to scientific notation Put decimal after first nonzero number 

4.2 Exponent is how many times decimal moved, 3 

x 10~ 3 Negative exponent, standard notation is small 

4.2 x 10 -3 Our Solution 



188 



Example 223. 



Convert3.21 x 10 5 to standard notation Positive exponent means standard notation 



321,000 



big number. Move decimal right 5 places 
Our Solution 



Example 224. 



Conver7.4x 10 3 to standard notation Negative exponent means standard notation 

is a small number. Move decimal left 3 places 
0.0074 Our Solution 



Converting between standard notation and scientific notation is important to 
understand how scientific notation works and what it does. Here our main 
interest is to be able to multiply and divide numbers in scientific notation using 
exponent properties. The way we do this is first do the operation with the front 
number (multiply or divide) then use exponent properties to simplify the 10's. 
Scientific notation is the only time where it will be allowed to have negative expo- 
nents in our final solution. The negative exponent simply informs us that we are 
dealing with small numbers. Consider the following examples. 



Example 225. 



(2.1xl0" 7 )(3.7xl0 5 ) 

(2.1)(3.7) = 7.77 

10- 7 10 5 = 10" 2 

7.77 xlO" 2 



Deal with numbers and 10's separately 
Multiply numbers 

Use product rule on 10's and add exponents 
Our Solution 



Example 226. 



4.96 x 10 4 



3.1 x 10" 3 
4.96 



3.1 
10 4 



io- 



Deal with numbers and 10's separately 



1.6 Divide Numbers 



10 7 Use quotient rule to subtract exponents, be careful with negatives! 



Be careful with negatives, 4— ( — 3) =4 + 3 = 7 
1.6 x 10 7 Our Solution 



189 



Example 227. 

(1.8 x 10~ 4 ) 3 Use power rule to deal with numbers and 10's separately 

1.8 3 = 5.832 Evaluate 1.8 3 

(10- 4 ) 3 = 10 -12 Multiply exponents 

5.832 xlO" 12 Our Solution 

Often when we multiply or divide in scientific notation the end result is not in sci- 
entific notation. We will then have to convert the front number into scientific 
notation and then combine the 10's using the product property of exponents and 
adding the exponents. This is shown in the following examples. 

Example 228. 

(4.7 x 10~ 3 )(6.1 x 10 9 ) Deal with numbers and 10's separately 

(4.7)(6.1) =28.67 Multiply numbers 

2.867 x 10 1 Convert this number into scientific notation 

10 1 10~ 3 10 9 = 10 7 Use product rule, add exponents, using 10 1 from conversion 

2.867 x 10 7 Our Solution 

World View Note: Archimedes (287 BC - 212 BC), the Greek mathematician, 
developed a system for representing large numbers using a system very similar to 
scientific notation. He used his system to calculate the number of grains of sand it 
would take to fill the universe. His conclusion was 10 63 grains of sand because he 
figured the universe to have a diameter of 10 14 stadia or about 2 light years. 

Example 229. 

2.014 x 10" 3 



3.8 xlO" 7 
2.014 



Deal with numbers and 10's separately 



3.8 



0.53 Divide numbers 



0.53 = 5.3x10 x Change this number into scientific notation 

10 — 1 10 -3 

— — = 10 3 Use product and quotient rule, using 10 _1 from the conversion 

Be careful with signs: 

(-l) + (-3)-(-7) = (-l) + (-3) + 7 = 3 
5.3 xlO 3 Our Solution 



190 



5.3 Practice - Scientific Notation 

Write each number in scientific notiation 

1) 885 2) 0.000744 

3) 0.081 4) 1.09 

5) 0.039 6) 15000 

Write each number in standard notation 

7) 8.7 x 10 5 8) 2.56 x 10 2 

9) 9 x 10" 4 10) 5 x 10 4 

11)2x10° 12)6xl0" 5 

Simplify. Write each answer in scientific notation. 



13 

15 
17 

19 

21 
23 

25 

27 

29 

31 
33 

35 
37 
39 
41 



(7 x 10" 1 )(2 x 10" 3 ) 14) (2 x 10" 6 )(8.8 x 10" 5 ) 

(5.26 x 10" 5 )(3.16 x lO" 2 ) 16) (5.1 x 10 6 )(9.84 x 10" 1 ) 

(2.6 x 10" 2 )(6x 10" 2 ) 

4.9 x 10 1 
2.7 x 10" 3 

5.33 x 10" 6 
9.62 x 10" 2 

(5.5 X 10" 5 ) 2 

(7.8 x 10" 2 ) 5 
(8.03 xlO 4 )" 4 

6.1 x 10" 6 
5.1 x 10" 4 

(3.6 x 10°)(6.1x 10" 3 ) 
(1.8 x 10" 5 )" 3 

9x 10 4 
7.83 x 10" 2 

3.22 x 10" 3 



7x 10" 6 

2.4 x 10" 6 
6.5 x 10° 

6x 10 3 
5.8 x 10" 3 



1Q x 7.4 xlO 4 
' 1.7 x lO" 4 




of) N 7.2 xlO" 1 
AV > 7.32 xlO" 1 




00 n 3.2 x 10" 3 
1 5.02 x 10° 




24) (9.6 x lO 3 )" 4 




26) (5.4 x 10 6 )" 3 




28) (6.88 x 10" 4 )(4. 


23 x 10 1 


30) i ax '°: 

1 7x 10~ 2 




32) (3.15xl0 3 )(8x 


10" 1 ) 


o,n 9.58 x 10" 2 
> 1.14 xlO" 3 




36) (8.3 x 10 1 ) 5 




-1Q\ 5xl ° 6 
°°^ 6.69 xlO 2 




40) (9x 10" 2 )" 3 




42) (2x 10 4 )(6x 10 


l ) 



191 



5.4 



Polynomials - Introduction to Polynomials 



Objective: Evaluate, add, and subtract polynomials. 

Many applications in mathematics have to do with what are called polynomials. 
Polynomials are made up of terms. Terms are a product of numbers and/or vari- 
ables. For example, 5x, 2y 2 , — 5, ab 3 c, and x are all terms. Terms are connected 
to each other by addition or subtraction. Expressions are often named based on 
the number of terms in them. A monomial has one term, such as 3x 2 . A bino- 
mial has two terms, such as a 2 — b 2 . A Trinomial has three terms, such as ax 2 + 
bx + c. The term polynomial means many terms. Monomials, binomials, trino- 
mials, and expressions with more terms all fall under the umbrella of "polyno- 
mials". 

If we know what the variable in a polynomial represents we can replace the vari- 
able with the number and evaluate the polynomial as shown in the following 
example. 

Example 230. 



2x 2 — 4x + 6 when x = — 4 

2(-4) 2 -4(-4)+6 

2(16) -4(- 4) +6 

32 + 16 + 6 

54 



Replace variable x with — 4 

Exponents first 

Multiplication (we can do all terms at once) 

Add 

Our Solution 



It is important to be careful with negative variables and exponents. Remember 
the exponent only effects the number it is physically attached to. This means — 
3 2 = — 9 because the exponent is only attached to the 3. Also, ( — 3) 2 = 9 because 
the exponent is attached to the parenthesis and effects everything inside. When 
we replace a variable with parenthesis like in the previous example, the substi- 
tuted value is in parenthesis. So the ( — 4) 2 = 16 in the example. However, con- 
sider the next example. 



Example 231. 




— x 2 + 2x + 6 when x = 3 


Replace variable x with 3 


-(3) 2 + 2(3) + 6 


Exponent only on the 3, not negative 


-9 + 2(3) + 6 


Multiply 


-9+6+6 


Add 


3 


Our Solution 



192 



World View Note: Ada Lovelace in 1842 described a Difference Engine that 
would be used to calu elate values of polynomials. Her work became the founda- 
tion for what would become the modern computer (the programming language 
Ada was named in her honor), more than 100 years after her death from cancer. 

Generally when working with polynomials we do not know the value of the vari- 
able, so we will try and simplify instead. The simplest operation with polynomials 
is addition. When adding polynomials we are mearly combining like terms. Con- 
sider the following example 



Example 232. 

(4a; 3 - 2x + 8) + (3a; 3 - 9x 2 - 11) Combine like terms 4a; 3 + 3a; 3 and 8-11 
7a; 3 — 9a; 2 — 2x — 3 Our Solution 



Generally final answers for polynomials are written so the exponent on the vari- 
able counts down. Example 3 demonstrates this with the exponent counting down 
3, 2, 1, (recall x° = 1). Subtracting polynomials is almost as fast. One extra step 
comes from the minus in front of the parenthesis. When we have a negative in 
front of parenthesis we distribute it through, changing the signs of everything 
inside. The same is done for the subtraction sign. 



Example 233. 

(5a; 2 — 2x + 7) — (3a; 2 + 6x — 4) Distribute negative through second part 

5a; 2 — 2x + 7 — 3a; 2 — 6x + 4 Combine like terms 5a; 2 — 3a; 3 , — 2x — 6x , and 7 + 4 

2a; 2 - 8x + 1 1 Our Solution 



Addition and subtraction can also be combined into the same problem as shown 
in this final example. 



Example 234. 

(2a; 2 — 4x + 3) + (5a; 2 — 6x + 1) — (x 2 — 9x + 8) Distribute negative through 

2a; 2 — 4x + 3 + 5a; 2 — 6x + 1 — x 2 + 9a; — 8 Combine like terms 

6a; 2 — x — 4 Our Solution 



193 



5.4 Practice - Introduction to Polynomials 

Simplify each expression. 

1) — a 3 — a 2 + 6a — 21 when a = — 4 

2) n 2 + 3n — 11 whenn = — 6 

3) n 3 - In 2 + 15n - 20 when n = 2 

4) n 3 - 9n 2 + 23n - 21 when n = 5 

5) — 5n 4 — lln 3 — 9n 2 — n — 5 when n = — 1 

6) x 4 — 5x 3 — x + 13 when x = 5 

7) x 2 + 9x + 23 when x = - 3 

8) -6x 3 + 41x 2 -32x + llwhenx = 6 

9) x 4 — 6x 3 + x 2 — 24 when x = 6 



10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
20 



m 4 + 8m 3 + 14m 2 + 13m + 5 when m = — 6 

(5p — 5p 4 ) — (8p — 8p 4 ) 

(7m 2 + 5m 3 ) — (6m, 3 — 5m 2 ) 

(3n 2 + n 3 )-(2n 3 -7n 2 ) 

(x 2 + 5x 3 ) + (7x 2 + 3x 3 ) 

(8n + n 4 ) - (3n - 4n 4 ) 

(3v A +l) + (5-v 4 ) 

(l + 5p 3 )-(l-8p 3 ) 

(6x 3 + 5x) - (8x + 6x 3 ) 

(5n 4 + 6n 3 ) + (8 - 3n 3 - 5n 4 ) 

(8x 2 + l)-(6-x 2 -x 4 ) 



194 



21) 
22) 
23) 
24) 
25) 
26) 
27) 
28) 
29) 
30) 
31) 
32) 
33) 
34) 
35) 
36) 
37) 
38) 
39) 
40) 
41) 
42) 



3 + 6 4 ) + (7 + 26 + 6 4 ) 
1 + 6r 2 ) + (6r 2 - 2 - 3r 4 ) 
8x 3 + l)-(5x 4 -6x 3 + 2) 
4n 4 + 6) - (4n - 1 - n 4 ) 
2a + 2a 4 ) -(3a 2 - 5a 4 + 4a) 
6v + 8v 3 ) + (3 + 4v 3 -3v) 
4p 2 - 3 - 2p) - (3p 2 - 6p + 3) 

7 + 4m + 8m 4 ) - (5m 4 + 1 + 6m) 
46 3 + 76 2 -3) + (8 + 56 2 + 6 3 ) 

7n + 1 - 8n 4 ) - (3n + 7n 4 + 7) 
3 + 2n 2 + 4n 4 ) + (n 3 - 7n 2 - 4n 4 ) 
7x 2 + 2x 4 + 7x 3 ) + (6x 3 - 8x 4 - 7a: 2 ) 
n — 5n 4 + 7) + (n 2 — 7n 4 — n) 
8x 2 + 2x 4 + 7x 3 ) + (7x 4 - 7x 3 + 2x 2 ) 

g r 4 _ 5r 3 + 5r 2) + ( 2r 2 + 2r 3 _ 7r 4 + X ) 

4x 3 + x - 7x 2 ) + (x 2 -8 + 2x + 6x 3 ) 
2n 2 + 7n 4 - 2) + (2 + 2n 3 + 4n 2 + 2n 4 ) 
U 3 - 46 + 46 4 ) - (86 3 - 46 2 + 26 4 - 86) 

8 - 6 + 76 3 ) - (36 4 + 76 - 8 + 76 2 ) + (3 - 36 + 66 3 ) 



l-3n 4 - 



in 



(7n 4 + 2 - 6n 2 + 3n 3 ) + (4n 3 + 8n 4 + 7) 



8x 4 + 2x 3 + 2x) + (2x + 2 - 2x 3 - x A ) - (x 3 + 5x 4 + 8x) 
6x - 5x 4 - Ax 2 ) - (2x - 7x 2 - Ax 4 - 8) - (8 - 6x 2 - 4x 4 ) 



195 



5.5 

Polynomials - Multiplying Polynomials 

Objective: Multiply polynomials. 

Multiplying polynomials can take several different forms based on what we are 
multiplying. We will first look at multiplying monomials, then monomials by 
polynomials and finish with polynomials by polynomials. 

Multiplying monomials is done by multiplying the numbers or coefficients and 
then adding the exponents on like factors. This is shown in the next example. 

Example 235. 

(4x 3 y A z) (2x 2 y 6 z 3 ) Multiply numbers and add exponents for x , y , and z 
8x 5 y 10 z 4 Our Solution 

In the previous example it is important to remember that the z has an exponent 
of 1 when no exponent is written. Thus for our answer the z has an exponent of 
1+3 = 4. Be very careful with exponents in polynomials. If we are adding or sub- 
tracting the exponnets will stay the same, but when we multiply (or divide) the 
exponents will be changing. 

Next we consider multiplying a monomial by a polynomial. We have seen this 
operation before with distributing through parenthesis. Here we will see the exact 
same process. 

Example 236. 

4x 3 (5x 2 — 2x + 5) Distribute the 4x 3 , multiplying numbers, adding exponents 
20x 5 -8x 4 + 20:e 3 Our Solution 

Following is another example with more variables. When distributing the expo- 
nents on a are added and the exponents on b are added. 

Example 237. 

2a 3 6(3ab 2 — 4a) Distribute, multiplying numbers and adding exponents 
6a 4 b 3 — 8a 4 b Our Solution 

There are several different methods for multiplying polynomials. All of which 
work, often students prefer the method they are first taught. Here three methods 
will be discussed. All three methods will be used to solve the same two multipli- 
cation problems. 

Multiply by Distributing 

196 



Just as we distribute a monomial through parenthesis we can distribute an entire 
polynomial. As we do this we take each term of the second polynomial and put it 
in front of the first polynomial. 

Example 238. 

(4a; + 7y) (3x — 2y) Distribute (4a; + 7y) through parenthesis 

3x(4x + 7y) — 2y(4x + 7y) Distribute the 3x and — 1y 

12a; 2 + 2\xy — 8xy — 14y 2 Combine like terms 2 lxy — 8xy 

12x 2 + 13xy - 14y 2 Our Solution 

This example illustrates an important point, the negative/subtraction sign stays 
with the 2y. Which means on the second step the negative is also distributed 
through the last set of parenthesis. 

Multiplying by distributing can easily be extended to problems with more terms. 
First distribute the front parenthesis onto each term, then distribute again! 

Example 239. 

(2x — 5) (4a; 2 — 7x + 3) Distribute (2x — 5) through parenthesis 

4x 2 (2x — 5) — 7a;(2:r — 5) + 3(2x — 5) Distribute again through each parenthesis 

8a; 3 — 20x 2 — 14a; 2 + 35a; + 6x — 15 Combine like terms 

8a; 3 - 34a; 2 + 41x - 15 Our Solution 

This process of multiplying by distributing can easily be reversed to do an impor- 
tant procedure known as factoring. Factoring will be addressed in a future lesson. 

Multiply by FOIL 

Another form of multiplying is known as FOIL. Using the FOIL method we mul- 
tiply each term in the first binomial by each term in the second binomial. The 
letters of FOIL help us remember every combination. F stands for First, we mul- 
tiply the first term of each binomial. O stand for Outside, we multiply the outside 
two terms. I stands for Inside, we multiply the inside two terms. L stands for 
Last, we multiply the last term of each binomial. This is shown in the next 
example: 



Example 240. 






(4x + 7y)(3x 


-2t/) 


Use FOIL to multiply 


(4x)(3x) = 


= 12a; 2 


F — First terms (Ax) (3a;) 


(4x)(-2y) = - 


- 8xy 


— Outside terms (4a;) ( — 2y) 


(7y)(3x) = 


-2lxy 


I — Inside terms (7y) (3x) 


(7y)(-2y) = - 


-Uy 2 


L — Last terms (7y) ( — 2y) 


12x 2 — 8xy + 21xy - 


-Uy 2 


Combine like terms —8xy + 2lxy 


12x 2 + 13xy- 


-Uy 2 


Our Solution 



197 



Some students like to think of the FOIL method as distributing the first term 4x 
through the (3a; — 2y) and distributing the second term ly through the (3x — 2y). 
Thinking about FOIL in this way makes it possible to extend this method to 
problems with more terms. 

Example 241. 

(2x — 5) (4x 2 — 7x + 3) Distribute 2x and — 5 

(2a;) (4a; 2 ) + (2a;) ( - 7x) + (2a;) (3) - 5(4a; 2 ) - 5( - 7x) - 5(3) Multiply out each term 

8x 3 — 14a; 2 + 6x — 20x 2 + 35x — 15 Combine like terms 

8a: 3 - 34a; 2 + 41a; - 15 Our Solution 

The second step of the FOIL method is often not written, for example, consider 
the previous example, a student will often go from the problem (4a; + ly)(3x — 2y) 
and do the multiplication mentally to come up with 12a; 2 — 8xy + 21xy — 14t/ 2 and 
then combine like terms to come up with the final solution. 

Multiplying in rows 

A third method for multiplying polynomials looks very similar to multiplying 
numbers. Consider the problem: 

35 
x27 

245 Multiply 7 by 5 then 3 

700 Use for placeholder, multiply 2 by 5 then 3 
945 Add to get Our Solution 

World View Note: The first known system that used place values comes from 
Chinese mathematics, dating back to 190 AD or earlier. 

The same process can be done with polynomials. Multiply each term on the 
bottom with each term on the top. 



Example 242. 




(4a; + ly) (3a; - 


-2y) 


4a: + 7y 


x 3a; 


-2y 


-8xy- 


Uy 2 


I2x 2 + 21xy 




12a; 2 + 13a; y- 


Uy 2 



Rewrite as vertical problem 



Multiply — 2y by 7ythen4x 

Multiply 3a; by ly then Ax. Line up like terms 

Add like terms to get Our Solution 



This same process is easily expanded to a problem with more terms. 



198 



Example 243. 






(2x- 


-5) (4a 


; 2 - 7x + 3) 




4a; 3 - 7x + 3 






x 2x 


-5 




-20x 2 


+ 35x- 


■15 


8x 3 - 


-14a; 2 


+ 6x 




8a; 3 - 


-34a; 2 


+ 41x- 


■15 



Rewrite as vertical problem 

Put polynomial with most terms on top 

Multiply — 5 by each term 

Multiply 2x by each term. Line up like terms 

Add like terms to get our solution 

This method of multiplying in rows also works with multiplying a monomial by a 
polynomial! 

Any of the three described methods work to multiply polynomials. It is suggested 
that you are very comfortable with at least one of these methods as you work 
through the practice problems. All three methods are shown side by side in the 
example. 

Example 244. 

(2x — y)(4x — by) 

Distribute FOIL Rows 

Ax(2x — y) — hy{2x — y) 2x(4x) + 2x( — 5y) — y(4x) — y( — 5y) 2x — y 

8x 2 — 4xy — lOxy — 5y 2 8x 2 — lOxy — 4xy + 5y 2 x 4x — by 

8x 2 — 14xy — 5y 2 8x 2 — 14xy + 5y 2 —I0xy + 5y 2 

8x 2 — 4xy 

8x 2 -14xy + 5y 2 

When we are multiplying a monomial by a polynomial by a polynomial we can 
solve by first multiplying the polynomials then distributing the coefficient last. 
This is shown in the last example. 

Example 245. 

3 (2x — 4)(x + 5) Multiply the binomials, we will use FOIL 

3(2x 2 + 10a; — 4a; — 20) Combine like terms 

3(2a; 2 + 6x-20) Distribute the 3 

6a; 2 + 18a; — 60 Our Solution 

A common error students do is distribute the three at the start into both paren- 
thesis. While we can distribute the 3 into the (2x — 4) factor, distributing into 
both would be wrong. Be careful of this error. This is why it is suggested to mul- 
tiply the binomials first, then distribute the coeffienct last. 



199 



5.5 Practice - Multiply Polynomials 



Find each product. 

l)6(p-7) 

3) 2(6x + 3) 
5) 5m 4 (4m + 4) 
7) (4n + 6)(8n + 8) 
9) (86 + 3)(7fe-5) 



11 
13 

15 
17 
19 
21 
23 
25 
27 
29 
31 
33 
35 
37 
39 



4x + 5)(2x + 3) 
3u-4)(5u-2) 
6x-7)(4x+l) 

5x + y)(6x — 4y) 
x + 3y)(3x + 4t/) 
7x + 5y)(8x + 3y) 



7)(6r 2 



5) 



6n-4)(2n 2 -2n + 5) 
6x + 3y) (6x 2 - 7x y + Ay 2 ) 
8n 2 + 4n + 6) (6n 2 - 5n + 6) 
5& 2 + 3£; + 3)(3£; 2 + 3£; + 6) 

3(3x-4)(2x + l) 

3(2x + l)(4a;-5) 

7(x-5)(x-2) 

6(4x-l)(4a: + l) 



2) 4k(8k + 4) 




4) 


3n 2 (6n + 7) 




6) 


3(4r-7) 




8) (2x + l)(x-4) 




10 


) (r + 8)(4r + 8) 




12 


) (7n-6)(n + 7) 




14 


) (6a + 4)(o-8) 




16 


) (5a:-6)(4x-l) 




18 


) (2u + 3v)(8u-7v) 




20 


) (8u + 6v)(5u-8v) 




22 


) (5a + 86) (a -36) 




24 


) (4s + 8)(4x 2 + 3x + 5) 




26 


) (2o-3)(46 2 + 46 + 4) 




28 


(3m — 2n) (7m 2 + 6m n 


+ 4n 2 ) 


30 


) (2a 2 + 6a + 3) (7a 2 -6c 


i+l) 


32 


) (7u 2 + 8uv-6v 2 )(6u 2 


-\-4uv + 3v 


34 


) 5(x-4)(2a:-3) 




36 


) 2(4x+l)(2x-6) 




38 


) 5(2x-l)(4x + l) 




40 


) 3(2x + 3)(6x + 9) 





200 



5.6 

Polynomials - Multiply Special Products 



Objective: Recognize and use special product rules of a sum and differ- 
ence and perfect squares to multiply polynomials. 

There are a few shortcuts that we can take when multiplying polynomials. If we 
can recognize them the shortcuts can help us arrive at the solution much quicker. 
These shortcuts will also be useful to us as our study of algebra continues. 

The first shortcut is often called a sum and a difference. A sum and a differ- 
ence is easily recognized as the numbers and variables are exactly the same, but 
the sign in the middle is different (one sum, one difference). To illustrate the 
shortcut consider the following example, multiplied by the distributing method. 



Example 246. 








(0 + 6) (0-6) 


Distribute (a + b) 




a(a + b) — b(a + b) 


Distribute a and — b 




a 2 + ab — ab — b 2 


Combine like terms ab 




a 2 -b 2 


Our Solution 



ab 



The important part of this example is the middle terms subtracted to zero. 
Rather than going through all this work, when we have a sum and a difference we 
will jump right to our solution by squaring the first term and squaring the last 
term, putting a subtraction between them. This is illustrated in the following 
example 



Example 247. 

(x — 5) (x + 5) Recognize sum and difference 

x 2 — 25 Square both, put subtraction between. Our Solution 



This is much quicker than going through the work of multiplying and combining 
like terms. Often students ask if they can just multiply out using another method 
and not learn the shortcut. These shortcuts are going to be very useful when we 
get to factoring polynomials, or reversing the multiplication process. For this 
reason it is very important to be able to recognize these shortcuts. More examples 
are shown here. 



201 



Example 248. 



(3x + 7) (3x — 7) Recognize sum and difference 

9x 2 — 49 Square both, put subtraction between. Our Solution 



Example 249. 

(2x — 6y) (2x + 6y) Recognize sum and difference 

4x 2 — 36y 2 Square both, put subtraction between. Our Solution 

It is interesting to note that while we can multiply and get an answer like a 2 — b 2 
(with subtraction), it is impossible to multiply real numbers and end up with a 
product such as a 2 + b 2 (with addition). 

Another shortcut used to multiply is known as a perfect square. These are easy 
to recognize as we will have a binomial with a 2 in the exponent. The following 
example illustrates multiplying a perfect square 

Example 250. 

(a + b) 2 Squared is same as multiplying by itself 

(a + b) (a + b) Distribute (a + b) 

a(a + b) + b(a + b) Distribute again through final parenthesis 

a 2 + ab + ab + b 2 Combine like terms ab + ab 

a 2 + 2ab + b 2 Our Solution 



This problem also helps us find our shortcut for multiplying. The first term in the 
answer is the square of the first term in the problem. The middle term is 2 times 
the first term times the second term. The last term is the square of the last term. 
This can be shortened to square the first, twice the product, square the last. If we 
can remember this shortcut we can square any binomial. This is illustrated in the 
following example 

Example 251. 



(x-5) 2 


Recognize perfect square 


x 2 


Square the first 


2(x)(-5) = -10x 


Twice the product 


(-5) 2 = 25 


Square the last 


x 2 -10x + 25 


Our Solution 



202 



Be very careful when we are squaring a binomial to NOT distribute the square 
through the parenthesis. A common error is to do the following: (x — 5) 2 = x 2 — 25 
(or x 2 + 25). Notice both of these are missing the middle term, — lOcc. This is 
why it is important to use the shortcut to help us find the correct solution. 
Another important observation is that the middle term in the solution always has 
the same sign as the middle term in the problem. This is illustrated in the next 
examples. 



Example 252. 



(2x + 5) 2 

(2x) 2 = 4x 2 

2(2x)(5)=20x 

5 2 = 25 

4x 2 + 20a; + 25 



Recognize perfect square 
Square the first 
Twice the product 
Square the last 
Our Solution 



Example 253. 



9x 2 



(Sx — 7y) 2 Recognize perfect square 
42xy + 49y 2 Square the first, twice the product, square the last. Our Solution 



Example 254. 



25a z 



(5a- 
90ab - 



- 96) 2 Recognize perfect square 
816 2 Square the first, twice the product, square the last. Our Solution 



These two formulas will be important to commit to memory. The more familiar 
we are with them, the easier factoring, or multiplying in reverse, will be. The final 
example covers both types of problems (two perfect squares, one positive, one 
negative), be sure to notice the difference between the examples and how each for- 
mula is used 

Example 255. 



(4x-7)(4x + 7) 
16x 2 -49 



(4x + 7) 2 
16a; 2 + 56x + 49 



(4x - 7) 2 
16x 2 -56x + 49 



World View Note: There are also formulas for higher powers of binomials as 



well, such as (a + b) c 



3a 2 b + 3a 6 2 + o 3 . While French mathematician Blaise 



Pascal often gets credit for working with these expansions of binomials in the 17th 
century, Chinese mathematicians had been working with them almost 400 years 
earlier! 



203 



5.6 Practice - Multiply Special Products 



Find each product. 

1) (x + 8)(x-8) 
3) (l + 3p)(l-3p) 

5) (l-7n)(l + 7n) 
7) (5n-8)(5n + 8) 
9) (4x + 8)(4x-8) 



11 
13 
15 
17 
19 
21 
23 
25 
27 
29 
31 
33 
35 
37 
39 



4y — x)(4y + x) 
4m — 8n) (4m + 8n) 
6x — 2y)(6x + 2y) 

a + 5) 2 

rc-8) 2 

p + 7) 2 

7-5n) 2 

5m -8) 2 

5x + 7y) 2 

2x + 2y) 2 

5 + 2r) 2 

2 + 5x) 2 

4v - 7) (4v + 7) 

n — 5)(n + 5) 

4& + 2) 2 



2)( 


a-4)(a + 4) 




4)( 


x — 3)(x + 3) 




6)( 


8m + 5) (8m — 


5) 


8) (2r + 3)(2r-3) 


10) 


(6-7)(6 + 7) 




12) 


(7a + 76) (7a - 


-76) 


14) 


(3y-3x)(3y- 


f 3x) 


16) 


(l + 5n) 2 




18) 


(v + 4) 2 




20) 


(l-6n) 2 




22) 


(7k -7) 2 




24) 


(4x-5) 2 




26) 


(3a + 36) 2 




28) 


(4m — n) 2 




30) 


(%x + hy) 2 




32) 


{rn-7) 2 




34) 


(8n + 7)(8n- 


7) 


36) 


(6 + 4)(6-4) 




38) 


(7x + 7) 2 




40) 


(3a -8) (3a + 


8) 



204 



5.7 



Polynomials - Divide Polynomials 



Objective: Divide polynomials using long division. 

Dividing polynomials is a process very similar to long division of whole numbers. 
But before we look at that, we will first want to be able to master dividing a 
polynomial by a monomial. The way we do this is very similar to distributing, 
but the operation we distribute is the division, dividing each term by the mono- 
mial and reducing the resulting expression. This is shown in the following exam- 
ples 



Example 256. 



9a; 5 + 6a; 4 - 18a; 3 - 24a; 2 
3a; 2 

9a; 5 6a; 4 18a; 3 24a; 2 
4 



3a: 2 3a; 2 3a; 2 3a; 2 



Divide each term in the numerator by 3a; 2 
Reduce each fraction, subtracting exponents 



3a; 3 + 2x 2 — 6a; — 8 Our Solution 



Example 257. 



ix 



4a; 2 - 2a; + 6 



ix s 4x 
4 



Ax 2 
- 2x 



6 



4a; 2 4a; 2 4x 2 4a; 2 



2a; + 1 



2a; 2a; 2 



Divide each term in the numerator by 4a; 2 

Reduce each fraction, subtracting exponents 
Remember negative exponents are moved to denominator 
Our Solution 



The previous example illustrates that sometimes we will have fractions in our 
solution, as long as they are reduced this will be correct for our solution. Also 

4x 2 

interesting in this problem is the second term —^ divided out completely. 
Remember that this means the reduced answer is 1 not 0. 

Long division is required when we divide by more than just a monomial. Long 
division with polynomials works very similar to long division with whole numbers. 



205 



An example is given to review the (general) steps that are used with whole num- 
bers that we will also use with polynomials 



Example 258. 

. 6 

4 631 Divide front numbers: — = 1... 

1 4 

1 

4 1 631 Multiply this number by divisor: 1-4 = 4 

— 4 Change the sign of this number (make it subtract) and combine 

23 Bring down next number 

23 
15 Repeat, divide front numbers: — = 5... 

4|63T 
-4 



23 Multiply this number by divisor: 5-4 = 20 
— 20 Change the sign of this number (make it subtract) and combine 
31 Bring down next number 



31 
157 Repeat, divide front numbers: — = 7... 

4|63T 

-4 

23 

-20 



31 Multiply this number by divisor: 7 • 4 = 28 

— 28 Change the sign of this number (make it subtract) and combine 

3 We will write our remainder as a fraction, over the divisor, added to the end 

3 

157— Our Solution 
4 



This same process will be used to multiply polynomials. The only difference is we 
will replace the word "number" with the word "term" 

Dividing Polynomials 

1. Divide front terms 

2. Multiply this term by the divisor 

206 



3. Change the sign of the terms and combine 

4. Bring down the next term 

5. Repeat 

Step number 3 tends to be the one that students skip, not changing the signs of 
the terms would be equivalent to adding instead of subtracting on long division 
with whole numbers. Be sure not to miss this step! This process is illustrated in 
the following two examples. 

Example 259. 

3x 3 - 5x 2 -32^ + 7 



x-4 



Rewrite problem as long division 



3x 3 



x — 4\3x — 5x — 32x + 7 Divide front terms: = 3x ^ 

x 



3x' 



x — 4|3x 3 — 5x 2 — 32x + 7 Multiply this term by divisor: 3x 2 (x — 4) = 3x 3 — \2x 2 

— 3a; 3 + 12a; 2 Change the signs and combine 

7a; 2 — 32a; Bring down the next term 

lx 2 

3x 2 + 7a: Repeat, divide front terms: = lx 

x 



x-4|3x 3 -5x 2 -32x + 7 
-3:c 3 + 12:c 2 



lx 2 — 32x Multiply this term by divisor: 7x(x — 4) = 7x 2 — 28a: 

7a: 2 + 28a: Change the signs and combine 

— 4a; + 7 Bring down the next term 



4x 

3x + lx — 4 Repeat, divide front terms: = — 4 

x 



x - 


-4|3x 3 -5x 2 -32x + 7 




-3x 3 + 12a; 2 




7x 2 - 32x 




-7x 2 + 28x 




-4x + 7 




+ 4a; - 16 



Ax + 7 Multiply this term by divisor: — 4(x — 4) = — 4x + 16 
6 Change the signs and combine 
9 Remainder put over divisor and subtracted (due to negative) 



207 



9 

3x 2 + 7x — 4 Our Solution 

x-4 



Example 260. 

6a; 3 - 8a; 2 + 10a; + 103 

2a; + 4 



Rewrite problem as long division 



6a; 3 



2x + 416a; 3 - 8a; 2 + 10a; + 103 Divide front terms: = 3a; 2 

2x 



3a; 2 



2x + 4|6x 3 - 8x 2 + 10a; + 103 Multiply term by divisor: 3x 2 (2a; + 4) = 6a; 3 + 12a; 2 

— 6a; 3 — 12a; 2 Change the signs and combine 

— 20a; 2 + 10a; Bring down the next term 

3x 2 - 10a; 

_ 20a; 2 

2x + 4|6x 3 — 8a; 2 + 10a; + 103 Repeat, divide front terms: — = — 10a; 

2x 

— 6a; 3 — 12a; 2 Multiply this term by divisor: 

- 20a; 2 + 10a; - 10a; (2a; + 4) = - 20a; 2 -40a; 

+ 20a; 2 + 40a; Change the signs and combine 

50a; + 103 Bring down the next term 

3a; 2 -10a; + 25 

5Q X 

2x + 4|6x 3 — 8a; 2 + 10a; + 103 Repeat, divide front terms: = 25 

2a; 



4|6x 3 
— 6a; 3 


-8a; 2 
-12a 


+ 10a; + 103 

2 




-20ar 
+ 20ar 


2 + 10x 
2 + 40x 






50x + 103 
50a; - 100 



Multiply this term by divisor: 25 (2x + 4) = 50a; + 100 

Change the signs and combine 

Remainder is put over divsor and added (due to positive) 



3a; 2 - 10a; + 25 + Our Solution 

2x + 4 

In both of the previous example the dividends had the exponents on our variable 
counting down, no exponent skipped, third power, second power, first power, zero 
power (remember x° = 1 so there is no variable on zero power). This is very 
important in long division, the variables must count down and no exponent can 
be skipped. If they don't count down we must put them in order. If an exponent 
is skipped we will have to add a term to the problem, with zero for its coefficient. 
This is demonstrated in the following example. 

208 



Example 261. 

2x 3 + 42 - Ax 

x + 3 



x 



x 



Reorder dividend, need x 2 term, add Ox 2 for this 



2 X 3 

x + 3 12a; 3 + Ox 2 — Ax + 42 Divide front terms: = 2x 2 



x 



2x' 



f 3|2x 3 + Ox 2 - Ax + 42 Multiply this term by divisor: 2x 2 (x + 3) = 2x 3 + 6x 2 

— 2x 3 — 6a; 2 Change the signs and combine 

— 6a; 2 — 4a; Bring down the next term 

2x 2 - 6x 

_ — Qx 

f 3|2x 3 + 0x 2 — Ax + 42 Repeat, divide front terms: = — 6x 

x 

— 2x 3 — 6x 2 



— 6x 2 — Ax Multiply this term by divisor: — 6x(x + 3) = — 6x 2 — 18a: 

+ 6a; 2 + 18a; Change the signs and combine 

14a; + 42 Bring down the next term 

2x 2 - 6x + 14 



14 x 

x + 3|2x 3 + 0a; 2 — Ax + 42 Repeat, divide front terms: = 14 

x 



c 3 + 0x 2 
2x 3 — 6x 2 



■ 6x 2 — Ax 
6x 2 + 18a: 

14a; + 42 Multiply this term by divisor: 14(x + 3) = 14x + 42 

— 14a; — 42 Change the signs and combine 

No remainder 

2a: 2 — 6a; + 14 Our Solution 



It is important to take a moment to check each problem to verify that the expo- 
nents count down and no exponent is skipped. If so we will have to adjust the 
problem. Also, this final example illustrates, just as in regular long division, 
sometimes we have no remainder in a problem. 

World View Note: Paolo Rufiini was an Italian Mathematician of the early 
19th century. In 1809 he was the first to describe a process called synthetic divi- 
sion which could also be used to divide polynomials. 



209 



5.7 Practice - Divide Polynomials 



Divide. 

-, x 20a; 4 + x 3 + 2x 2 



V 


4x 3 


3) 


20n 4 + n 3 + 40n 2 


lOn 


5) 


12a: 4 + 24a: 3 + 3x 2 


6x 


7) 


10n 4 + 50n 3 + 2n 2 


10n 2 


9) 


a: 2 - 2a: - 71 
a: + 8 


11) 


n 2 + 13n + 32 
n + 5 


13) 


« 2 - 2ti - 89 
u-10 


15) 


a 2 - 4a - 38 
a-8 


17) 


45p 2 + 56p+19 


9p + 4 


19) 


10a: 2 - 32x + 9 


10a; -2 


21) 


4 r 2 _ j, _ i 
4r + 3 


23) 


n 2 -4 
n-2 


25) 


276 2 + 876 + 35 


36 + 8 


27) 


4a: 2 - 33a: + 28 


Ax — 5 


29) 


a 3 + 15a 2 + 49a - 55 


a + 7 


31) 


x 3 - 26a; - 41 
x + 4 


33) 


3n 3 + 9n 2 - 64n - 68 


n + 6 


35) 


x 3 - 46a; + 22 

x + 7 


37) 


9p 3 + 45p 2 + 27p - 5 


9p + 9 


39) 


r 3 -r 2 - 16r + 8 


r — 4 


41) 


12n 3 + 12n 2 - 15n - 4 


2n + 3 


A1\ 


4i) 3 - 21t> 2 + 6u + 19 



^ 5a: 4 + 45a: 3 + 4a; 2 


^J 


9x 


^ 3fc 3 + 4fc 2 + 2fc 


4 J 


8k 


^ 5p 4 + 16p 3 + 16p 2 


DJ 


4p 


Q ^ 3m 4 + 18m 3 + 27m 2 


°) 


9m 2 


10) 


r 2 - 3r - 53 
r-9 


12) 


b 2 - 106 + 16 

6-7 


14) 


x 2 + 4x - 26 

x + 7 


16) 


x 2 - 10a: + 22 
a;- 4 


18) 


48fc 2 - 70A; + 16 


6fc-2 


20) 


n 2 + 7n + 15 
n + 4 


22) 


3m 2 + 9m - 9 


3m — 3 


24) 


2a; 2 - 5a; - 8 
2a; + 3 


26) 


3v 2 - 32 
3v-9 


28) 


4n 2 - 23n - 38 


4n + 5 


30) 


8fc 3 - 66fc 2 + 12fc + 37 


fc-8 


32) 


a; 3 - 16a; 2 + 71a; - 56 


a;-8 


34) 


fc 3 - 4fc 2 - 6fc + 4 


k- 1 


36) 


2n 3 + 21n 2 + 25n 


2n + 3 


38) 


8m 3 - 57m 2 + 42 


8m + 7 


■id) 


2a; 3 + 12a; 2 + 4a; - 37 




2a- + 6 


AO\ 


246 3 - 386 2 + 296 - 60 



46-7 



4^ + 3 



210 



Chapter 6 : Factoring 

6.1 Greatest Common Factor 212 

6.2 Grouping 216 

6.3 Trinomials where a =1 221 

6.4 Trinomials where a ^ 1 226 

6.5 Factoring Special Products 229 

6.6 Factoring Strategy 234 

6.7 Solve by Factoring 237 



211 



6.1 

Factoring - Greatest Common Factor 



Objective: Find the greatest common factor of a polynomial and factor 
it out of the expression. 

The opposite of multiplying polynomials together is factoring polynomials. There 
are many benifits of a polynomial being factored. We use factored polynomials to 
help us solve equations, learn behaviors of graphs, work with fractions and more. 
Because so many concepts in algebra depend on us being able to factor polyno- 
mials it is very important to have very strong factoring skills. 

In this lesson we will focus on factoring using the greatest common factor or GCF 
of a polynomial. When we multiplied polynomials, we multiplied monomials by 
polynomials by distributing, solving problems such as 4x 2 (2x 2 — 3a; + 8) = 8x 4 — 
12a; 3 + 32a;. In this lesson we will work the same problem backwards. We will 
start with 8a; 2 — 12a; 3 + 32a; and try and work backwards to the 4x 2 (2a; — 3x + 8). 

To do this we have to be able to first identify what is the GCF of a polynomial. 
We will first introduce this by looking at finding the GCF of several numbers. To 
find a GCF of sevearal numbers we are looking for the largest number that can be 
divided by each of the numbers. This can often be done with quick mental math 
and it is shown in the following example 



Example 262. 



Find the GCF of 15, 24, and 27 

15 24 27 

— = 5, — = 6, — = 9 Each of the numbers can be divided by 3 

GCF = 3 Our Solution 



When there are variables in our problem we can first find the GCF of the num- 



212 



bers using mental math, then we take any variables that are in common with each 
term, using the lowest exponent. This is shown in the next example 



Example 263. 



GCF of 24x V^ , 18a; V, and Ylx A yz 



24 
~6~ 



= 4, 



18 



3, 



12 



x y 



2 Each number can be divided by 6 

x and y are in all 3, using lowest exponets 



GCF = 6a: 2 ?/ Our Solution 



To factor out a GCF from a polynomial we first need to identify the GCF of all 
the terms, this is the part that goes in front of the parenthesis, then we divide 
each term by the GCF, the answer is what is left inside the parenthesis. This is 
shown in the following examples 



Example 264. 



4x 2 



- sy>^ 



Ax 2 — 20a; + 16 GCF is 4, divide each term by 4 
20a: „ 16 



5x, ^f- = 4 This is what is left inside the parenthesis 
Our Solution 



4 ' 4 

4(a; 2 -5a; + 4) 



With factoring we can always check our solutions by multiplying (distributing in 
this case) out the answer and the solution should be the original equation. 



Example 265. 



25a; 4 



5x z 



5x z 



25a: 4 
15a; 3 



3x 



15a; 3 + 20a; 2 GCF is 5a; 2 , divide each term by this 
20a; 2 



, ;4 This is what is left inside the parenthesis 

5a; 5ar 

5a; 2 (5a; 2 — 3a: + 4) Our Solution 



Example 266. 



3x y z + 5x y z — Axy GCF is a; y, divide each term by this 



213 



x y z = 3 x 2 z, y „ z = 5x 3 yz 5 , -%— = - 4y 2 This is what is left in parenthesis 

xy z xy z xy z 

xy 2 (3x 2 z + 5x 3 yz 5 — 4y 2 ) Our Solution 



World View Note: The first recorded algorithm for finding the greatest 
common factor comes from Greek mathematician Euclid around the year 300 BC! 



Example 267. 



21a; + 14a; + 7a; GCF is 7a;, divide each term by this 



21a; 3 9 14a; 2 7x 

= 3x , = 2x, — = 1 This is what is left inside the parenthesis 

7x 7x 7x 

7x{?>x 2 + 2x + 1) Our Solution 



It is important to note in the previous example, that when the GCF was 7x and 
7x was one of the terms, dividing gave an answer of 1. Students often try to 
factor out the 7x and get zero which is incorrect, factoring will never make terms 
dissapear. Anything divided by itself is 1, be sure to not forget to put the 1 into 
the solution. 

Often the second line is not shown in the work of factoring the GCF. We can 
simply identify the GCF and put it in front of the parenthesis as shown in the fol- 
lowing two examples. 



Example 


268. 


12a; V " 


- 6x 4 y A + 8x 3 y 5 


2a; 3 y 2 (6a; 


2 -3xy 2 + 4y 3 ) 


Example 


269. 



GCF is 2x 3 y 2 , put this in front of parenthesis and divide 
Our Solution 



18a 4 b 3 - 27a 3 6 3 + 9a 2 6 3 GCF is 9a 2 6 3 , divide each term by this 
9a 2 o 3 (2a 2 -3a + l) Our Solution 



Again, in the previous problem, when dividing 9a 2 o 3 by itself, the answer is 1, not 
zero. Be very careful that each term is accounted for in your final solution. 



214 



6.1 Practice - Greatest Common Factor 

Factor the common factor out of each expression. 

1)9 + 8b 2 2)x-5 

3)452 2 -25 4)l + 2n 2 

5)56-35p 6)50^-802/ 

7)7ab-35a 2 6 8)27*V-72*V 

9)-3a 2 b + 6a% 2 10)8^ + 4^ 

11) — 52 2 — 5x 3 — 15x 4 

13) 202 4 - 302 + 30 

15) 28m 4 + 40m 3 + 8 

17) 306 9 + 5a6-15a 2 

19) -48a 2 6 2 -56a 3 6-56a 5 6 

21) 20x 8 y 2 z 2 + 15x 5 y 2 z + 35x 3 y 3 z 



12) -32n 9 + 32n 6 + 40n 5 

14) 21p 6 + 30p 2 + 27 

16) -102 4 + 202 2 + 122 

18) 27y 7 +12|/ 2 2 + 92/ 2 

20) 30m 6 + 15mn 2 -25 

22) 3p + 12g-15g 2 r 2 

24) 30y 4 z 3 x 5 + 50y 4 z 5 -l0y 4 z 3 x 
23) 502 2 2/+ 10 2/ 2 + 702, 2 26) 286 + 14 , 2 + 35fo 3 + 76 5 



25) 30gpr - 5qp + 5q 2g) 3^3 + 6q5 + 27fl 3 + 21fl 2 

27) -18n 5 + 3n 3 -21n + 3 30 ) - 24x 6 - 4x 4 + 12x 3 + 4x 2 

29) -402 n -202 12 + 502 13 -502 14 32) - 10y 7 + 6y 10 - Ay 10 x - 8y 8 x 
31) — 32mn 8 + 4m 6 n + 12mn 4 + 16mn 



215 



Example 


271. 






3a(2 


a + 56) - 


- 76(2a 


+ 56) 




(2a + 


56) (3a 


-76) 



6.2 

Factoring - Grouping 

Objective: Factor polynomials with four terms using grouping. 

The first thing we will always do when factoring is try to factor out a GCF. This 
GCF is often a monomial like in the problem 5x y + lOxz the GCF is the mono- 
mial 5x, so we would have 5x(y + 1z). However, a GCF does not have to be a 
monomial, it could be a binomial. To see this, consider the following two example. 

Example 270. 

3ax — 7bx Both have x in common, factor it out 
x(3a — 76) Our Solution 

Now the same problem, but instead of x we have (2a + 56). 



Both have (2a + 56) in common, factor it out 
Our Solution 



In the same way we factored out a GCF of x we can factor out a GCF which is a 
binomial, (2a + 56). This process can be extended to factor problems where there 
is no GCF to factor out, or after the GCF is factored out, there is more factoring 
that can be done. Here we will have to use another strategy to factor. We will use 
a process known as grouping. Grouping is how we will factor if there are four 
terms in the problem. Remember, factoring is like multiplying in reverse, so first 
we will look at a multiplication problem and then try to reverse the process. 

Example 272. 

(2a + 3) (56 + 2) Distribute (2a + 3) into second parenthesis 
56(2a + 3) + 2(2a + 3) Distribute each monomial 
lOab + 156 + 4a + 6 Our Solution 

The solution has four terms in it. We arrived at the solution by looking at the 
two parts, 56(2a + 3) and 2 (2a + 3). When we are factoring by grouping we will 
always divide the problem into two parts, the first two terms and the last two 
terms. Then we can factor the GCF out of both the left and right sides. When we 
do this our hope is what is left in the parenthesis will match on both the left and 
right. If they match we can pull this matching GCF out front, putting the rest in 
parenthesis and we will be factored. The next example is the same problem 
worked backwards, factoring instead of multiplying. 

216 



Example 273. 



10ab + 156 + 4a + 6 


10ab + 156 +4a + 6 


56(2a + 3) +2(2o + 3) 



(2a + 3)(56 + 2) 



Split problem into two groups 
GCFonleft is 56, on the right is 2 
(2a + 3) is the same! Factor out this GCF 
Our Solution 



The key for grouping to work is after the GCF is factored out of the left and 
right, the two binomials must match exactly. If there is any difference between 
the two we either have to do some adjusting or it can't be factored using the 
grouping method. Consider the following example. 



Example 274. 



6x 2 + 9xy — 14a; — 21y 



6x 2 + 9xy — 


14a;- 


-21y 


3x(2x + 3y) + 7( 


-2x- 


-3y) 



Split problem into two groups 
GCF on left is 3x , on right is 7 
The signs in the parenthesis don't match! 



when the signs don't match on both terms we can easily make them match by fac- 
toring the opposite of the GCF on the right side. Instead of 7 we will use — 7. 
This will change the signs inside the second parenthesis. 



3x(2x + 3y) -7(2x + 3y) 



(2x + 3y)(3x-7) 



(2x + 3y) is the same! Factor out this GCF 
Our Solution 



Often we can recognize early that we need to use the opposite of the GCF when 
factoring. If the first term of the first binomial is positive in the problem, we will 
also want the first term of the second binomial to be positive. If it is negative 
then we will use the opposite of the GCF to be sure they match. 



Example 275. 



5xy — 8x — lOy + 16 Split the problem into two groups 



5xy — 8x — 10y + 16 GCFonleft is x, on right we need a negative, 

so we use — 2 



x(5y-8) -2(5?/ -8) (5y - 8) is the same! Factor out this GCF 



{by — 8) (x — 2) Our Solution 



217 



Sometimes when factoring the GCF out of the left or right side there is no GCF 
to factor out. In this case we will use either the GCF of 1 or — 1. Often this is all 
we need to be sure the two binomials match. 



Example 276. 



12c 


,6- 


14a -66 + 7 


12ab 


-14a -66 + 7 


2a(66- 


7) 


-1(66-7) 



(66-7)(2a-l) 



Split the problem into two groups 
GCF on left is 2a , on right, no GCF, use — 1 
(66 — 7) is the same! Factor out this GCF 
Our Solution 



Example 277. 



6x 3 — 15x 2 + 2a; — 5 



6x — 15x 2 



2x-5 



3x 2 (2x-5) +l(2x-5) 



(2x-5)(3x 2 + l) 



Split problem into two groups 
GCF on left is 3a; 2 , on right, no GCF, use 1 
(2x — 5) is the same! Factor out this GCF 
Our Solution 



Another problem that may come up with grouping is after factoring out the GCF 
on the left and right, the binomials don't match, more than just the signs are dif- 
ferent. In this case we may have to adjust the problem slightly. One way to do 
this is to change the order of the terms and try again. To do this we will move 
the second term to the end of the problem and see if that helps us use grouping. 



Example 278. 



4a 2 -216 3 + 6a6-14a6 2 



4a 2 


-216 3 


+ 6a6- 


14a6 2 


l(4a 2 - 


-216 3 ) 


+ 2a6(3 


-76) 


4a 2 + 6a 


b-Uab 2 


-216 3 


4a 2 


+ 6ab 


-14a6 2 - 


-216 3 


2a(2a + 36) 


-76 2 (2a + 36) 



(2a + 36) (2a - 76 2 



Split the problem into two groups 

GCF on left is 1, on right is 2a 6 

Binomials don't match! Move second term to end 

Start over, split the problem into two groups 

GCF on left is 2a , on right is - 76 2 

(2a + 36) is the same! Factor out this GCF 

Our Solution 



When rearranging terms the problem can still be out of order. Sometimes after 
factoring out the GCF the terms are backwards. There are two ways that this can 
happen, one with addition, one with subtraction. If it happens with addition, for 



218 



example the binomials are (a + b) and (b + a), we don't have to do any extra 
work. This is because addition is the same in either order (5 + 3 = 3 + 5 = 8). 



Example 279. 



7 + y 


— 3xy — 2\x 


7+y 


— 3xy — 2\x 


1(7 + ?/) 


-3x(y + 7) 



(y + 7)(l-3x) 



Split the problem into two groups 

GCF on left is 1 , on the right is — 3x 

y + 7 and 7 + y are the same, use either one 

Our Solution 



However, if the binomial has subtraction, then we need to be a bit more careful. 
For example, if the binomials are (o — b) and (b — a), we will factor out the oppo- 
site of the GCF on one part, usually the second. Notice what happens when we 
factor out — 1. 



Example 280. 



(b — a) Factor out — 1 
1 ( — b + a) Addition can be in either order, switch order 
— 1 (a — b) The order of the subtraction has been switched ! 



Generally we won't show all the above steps, we will simply factor out the oppo- 
site of the GCF and switch the order of the subtraction to make it match the 
other binomial. 



Example 281. 



8xy 


-12?/ + 15- 


-lOx 


8xy 


-12y 15- 


lOx 


4y(2x- 


3) +5(3- 


-2x) 


4y(2y- 


3) -5(2x 


-3) 



(2z-3)(4j/-5) 



Split the problem into two groups 

GCF on left is 4y, on right, 5 

Need to switch subtraction order, use — 5 in middle 

Now 2x — 3 match on both! Factor out this GCF 

Our Solution 



World View Note: Sofia Kovalevskaya of Russia was the first woman on the 
editorial staff of a mathematical journal in the late 19th century. She also did 
research on how the rings of Saturn rotated. 



219 



6.2 Practice - Grouping 



Factor each completely. 

I) 40r 3 -8r 2 -25r + 5 
3) 3n 3 - 2n 2 - 9n + 6 

5) 156 3 + 21fc 2 -356-49 
7) 3x 3 + 15x 2 + 2x + 10 
9) 35x 3 -28x 2 - 20x + 16 

II) 7xy-49x + 5y-35 

13) 32xy + 40a; 2 + 12y + 15x 
15) 16xy — 56x + 2y — 7 
17) 2xy - 8x 2 + 7m 3 - 28y 2 x 
19) 40icm + 35:e-8m 2 -7?/ 
21) 32uv-20u + 24v-l5 
23) 10icm + 30 + 25:e + 12m 
25) 3mv + 14u - 6m 2 - 7v 
27) 16xy-3x-6x 2 + 8y 



2) 35z 3 - 10x 2 - 56a; + 16 

4) 14t; 3 + 10t; 2 -7t;-5 
6) 6x 3 - 48x 2 + 5x - 40 
8) 28p 3 + 21p 2 + 20p + 15 
10) 7n 3 + 21n 2 -5n-15 
12) 42r 3 -49r 2 + 18r-21 
14) 15afe-6a + 56 3 -26 2 
16) 3mn — 8m + 15n — 40 
18) 5mn + 2m - 25n - 10 
20) 8xy + 56x-y-7 
22) Auv + 14m 2 + 12v + 42m 
24) 24xy + 25m 2 - 20s - 30m 3 
26) 56a6 + 14-49a-166 



220 



6.3 

Factoring - Trinomials where a = 1 

Objective: Factor trinomials where the coefficient of x 2 is one. 

Factoring with three terms, or trinomials, is the most important type of factoring 
to be able to master. As factoring is multiplication backwards we will start with a 
multipication problem and look at how we can reverse the process. 

Example 282. 

(x + 6) (x — 4) Distribute (x + 6) through second parenthesis 

x(x + 6) — 4(x + 6) Distribute each monomial through parenthesis 

x 2 + 6x — 4x — 24 Combine like terms 

x 2 + 2x - 24 Our Solution 



You may notice that if you reverse the last three steps the process looks like 
grouping. This is because it is grouping! The GCF of the left two terms is x and 
the GCF of the second two terms is — 4. The way we will factor trinomials is to 
make them into a polynomial with four terms and then factor by grouping. This 
is shown in the following example, the same problem worked backwards 



Example 283. 

x 2 + 2x — 24 Split middle term into + 6x — 4x 

x 2 + 6x — 4x — 24 Grouping: GCF on left is x , on right is — 4 

x(x + 6) — 4(x + 6) (x + 6) is the same, factor out this GCF 

(x + 6) (x — 4) Our Solution 



The trick to make these problems work is how we split the middle term. Why did 
we pick + 6x — 4x and not + 5x — 3x? The reason is because 6x — 4x is the only 
combination that works! So how do we know what is the one combination that 
works? To find the correct way to split the middle term we will use what is called 
the ac method. In the next lesson we will discuss why it is called the ac method. 
The way the ac method works is we find a pair of numers that multiply to a cer- 
tain number and add to another number. Here we will try to multiply to get the 
last term and add to get the coefficient of the middle term. In the previous 

221 



example that would mean we wanted to multiply to - 
only numbers that can do this are 6 and — 4 (6 • — 4 
This process is shown in the next few examples 



24 and add to 
- 24 and 6 + ( 



- 2. The 

4) =2). 



Example 


284. 






x 2 + 9^ + 18 


Want to multiply to 18, add to 9 




x 2 + 6x + 3x + 18 


6 and 3, split the middle term 




x(x + 6) + 3(x + 6) 


Factor by grouping 




(x + 6)(x + 3) 


Our Solution 


Example 


285. 






x 2 - 4x + 3 


Want to multiply to 3, add to — 4 




JL OiX/ JU | O 


— 3 and — 1 , split the middle term 




x(x — 3) — l(x — 3) 


Factor by grouping 




(x — 3)(x — 1) 


Our Solution 


Example 


286. 






x 2 -8x-20 


Want to multiply to — 20, add to - 




x 2 - lOx + 2x - 20 


— 10 and 2, split the middle term 




x(x-10) + 2(x-10) 


Factor by grouping 



(x-10)(x + 2) Our Solution 



Often when factoring we have two variables. These problems solve just like prob- 
lems with one variable, using the coefficients to decide how to split the middle 
term 



Example 287. 



a 2 -9ab + Ub 2 
a 2 -7ab-2ab + Ub 2 
a(a - 7b) - 2b(a - 7b) 

(a-7b)(a-2b) 



Want to multiply to 14, add to — 9 
— 7 and — 2, split the middle term 
Factor by grouping 
Our Solution 



222 



As the past few examples illustrate, it is very important to be aware of negatives 
as we find the pair of numbers we will use to split the middle term. Consier the 
following example, done incorrectly, ignoring negative signs 



Warning 


288. 










x 2 + 5x ■ 

x 2 + 2x + 3x - 

x(x + 2) + 3(a; - 


-6 
-6 

-2) 



Want to multiply to 6, add 5 
2 and 3, split the middle term 
Factor by grouping 
??? Binomials do not match! 



Because we did not use the negative sign with the six to find our pair of numbers, 
the binomials did not match and grouping was not able to work at the end. Now 
the problem will be done correctly. 



Example 289. 




x 2 + 5x — 6 


Want to multiply to — 6, add to 5 


x 2 + 6x — x — 6 


6 and — 1 , split the middle term 


x(x + 6) — l(x + 6) 


Factor by grouping 


(x + 6)(x-l) 


Our Solution 



You may have noticed a shortcut for factoring these problems. Once we identify 
the two numbers that are used to split the middle term, these are the two num- 
bers in our factors! In the previous example, the numbers used to split the middle 
term were 6 and — 1, our factors turned out to be (x + 6) (x — 1). This pattern 
does not always work, so be careful getting in the habit of using it. We can use it 
however, when we have no number (technically we have a 1) in front of x 2 . In all 
the problems we have factored in this lesson there is no number in front of x 2 . If 
this is the case then we can use this shortcut. This is shown in the next few 
examples. 



Example 290. 

x 2 — 7x — 18 Want to multiply to — 18, add to — 7 



— 9 and 2, write the factors 
(x — 9) (x + 2) Our Solution 



223 



Example 291. 

m 2 — mn — 30n 2 Want to multiply to — 30, add to — 1 

5 and — 6, write the factors, don't forget second variable 
(m + 5n) (m — 6n) Our Solution 



It is possible to have a problem that does not factor. If there is no combination of 
numbers that multiplies and adds to the correct numbers, then we say we cannot 
factor the polynomial, or we say the polynomial is prime. This is shown in the fol- 
lowing example. 



Example 292. 

x 2 + 2x + Q Want to multiply to 6, add to 2 

1 • 6 and 2 • 3 Only possibilities to multiply to six, none add to 2 

Prime, can't factor Our Solution 



When factoring it is important not to forget about the GCF. If all the terms in a 
problem have a common factor we will want to first factor out the GCF before we 
factor using any other method. 



Example 


293 


• 








Sx 2 - 


- 24x + 45 






3(x 2 - 


-8a; + 15) 






3(x - 


-5)(x-3) 



GCF of all terms is 3, factor this out 
Want to multiply to 15, add to — 8 
— 5 and — 3, write the factors 
Our Solution 



Again it is important to comment on the shortcut of jumping right to the factors, 
this only works if there is no coefficient on x 2 . In the next lesson we will look at 
how this process changes slightly when we have a number in front of x 2 . Be 
careful not to use this shortcut on all factoring problems! 

World View Note: The first person to use letters for unknown values was Fran- 
cois Vieta in 1591 in France. He used vowels to represent variables we are solving 
for, just as codes used letters to represent an unknown message. 



224 



6.3 Practice - Trinomials where a 



Factor each completely. 




1) p 2 + 17p + 72 


2) x 2 + x-12 


3) n 2 - 9n + 8 


4) x 2 + x-30 


5) x 2 -9a;-10 


6) x 2 + 13x + 40 


7) o 2 + 12o + 32 


8) b 2 - 17b + 70 


9) x 2 + 3x - 70 


10) x 2 + 3x-18 


11) n 2 -8n + 15 


12) a 2 -6a -27 


13) p 2 + 15p + 54 


14) p 2 + 7p-30 


15) n 2 — 15n + 56 


16) m 2 — 15m n + 50n 2 




18) m 2 — 3mn — 40n 2 


17) u 2 -8uv + 15v 2 






20) x 2 + 10xy + 16y 2 


19) m 2 + 2mn — 8n 2 






22) u 2 -9uv + Uv 2 


21) x 2 -llxy + 18y 2 


24) x 2 + lAxy + 4:5y 2 


23) x 2 + xt/-12t/ 2 


26) 4x 2 + 52^ + 168 


25) x 2 + Axy - \2y 2 


28) 5n 2 -45n + 40 


27) 5a 2 + 60a + 100 


30) 5t; 2 + 20t>-25 


29) 6a 2 + 24a -192 


32) 5m 2 + 30mn-90n 2 


31) 6x 2 + 18xy + 12y 2 


34) 6m 2 — 36mn — 162n 2 


33) 6x 2 + 96xy + 378y 2 





225 



6.4 

Factoring - Trinomials where a ^ 1 



Objective: Factor trinomials using the ac method when the coefficient 
of x 2 is not one. 

When factoring trinomials we used the ac method to split the middle term and 
then factor by grouping. The ac method gets it's name from the general trinomial 
equation, ax 2 + bx + c, where a, b, and c are the numbers in front of x 2 , 
x and the constant at the end respectively. 

World View Note: It was French philosopher Rene Descartes who first used let- 
ters from the beginning of the alphabet to represent values we know (a, b, c) and 
letters from the end to represent letters we don't know and are solving for (x, y, 
z). 

The ac method is named ac because we multiply a ■ c to find out what we want to 
multiply to. In the previous lesson we always multiplied to just c because there 
was no number in front of x 2 . This meant the number was 1 and we were multi- 
plying to lc or just c. Now we will have a number in front of x 2 so we will be 
looking for numbers that multiply to ac and add to b. Other than this, the pro- 
cess will be the same. 



Example 294. 

Sx 2 + llx + 6 Multiply to acor (3)(6) = 18, add to 11 

3x 2 + 9x + 2x + 6 The numbers are 9 and 2 , split the middle term 

3x(x + 3) + 2(x + 3) Factor by grouping 

(x + 3) (3x + 2) Our Solution 



When a = 1, or no coefficient in front of x 2 , we were able to use a shortcut, using 
the numbers that split the middle term in the factors. The previous example illus- 
trates an important point, the shortcut does not work when a =£ 1. We must go 
through all the steps of grouping in order to factor the problem. 



Example 


295. 








8.7. 


■ 2 -2x- 


-15 


8x 2 - 


-12a; 


+ 10x- 


-15 


4x(2x 


-3) 


+ 5(2x- 


-3) 




(2a;- 


-3)(4x + 5) 



Multiply to acor (8) ( - 15) = - 120, add to - 2 
The numbers are — 12 and 10, split the middle term 
Factor by grouping 
Our Solution 



226 



Example 296. 

10x 2 - 27a; + 5 Multiply to acor (10) (5) = 50, add to - 27 

10a; 2 — 25a; — 2x + 5 The numbers are — 25 and — 2, split the middle term 

5x(2x — 5) — l(2a; — 5) Factor by grouping 

(2x — 5) (5x — 1) Our Solution 

The same process works with two variables in the problem 

Example 297. 

Ax 2 -xy- 5y 2 Multiply to ocor (4)( - 5) = - 20, add to - 1 

4a: 2 + 4xy — 5xy — 5y 2 The numbers are 4 and — 5, split the middle term 

4x(x + y) — 5y(x + y) Factor by grouping 

(x + y) (4x — 5y) Our Solution 

As always, when factoring we will first look for a GCF before using any other 
method, including the ac method. Factoring out the GCF first also has the added 
bonus of making the numbers smaller so the ac method becomes easier. 



Example 298. 






18x 3 + 33x 2 - 


30a: 


3x[6x 2 + llx- 


-10] 


3a:[6a: 2 + 15a: 


-4a;- 


-10] 


3x[3x(2a; + 5) - 


■2(2x + 5)] 


3a;(2a; + 


5) (3a; 


-2) 



GCF = 3a; , factor this out first 

Multiply to acor (6)( - 10) = - 60, add to 11 

The numbers are 15 and — 4, split the middle term 

Factor by grouping 

Our Solution 



As was the case with trinomials when a = 1, not all trinomials can be factored. If 
there is no combinations that multiply and add correctly then we can say the tri- 
nomial is prime and cannot be factored. 

Example 299. 

3x 2 + 2a: - 7 Multiply to ac or (3) ( - 7) = - 21 , add to 2 
— 3(7) and — 7(3) Only two ways to multiply to — 21, it doesn't add to 2 
Prime, cannot be factored Our Solution 



227 



6.4 Practice - Trinomials where a ^ 1 



Factor each completely. 






1) 7x 2 -48x + 36 


2) 


7n 2 - 44n + 12 


3) 7b 2 + 15b + 2 


4) 


7v 2 -24v-16 


5) 5a 2 -13a -28 


6) 


5n 2 - 4n - 20 


7) 2x 2 - 5x + 2 


8) 


3r 2 - 4r - 4 


9) 2x 2 + 19^ + 35 


10 


) 7x 2 + 29a; - 30 


11) 2b 2 -b -3 


12 


) 5£; 2 -26£; + 24 


13) 5k 2 + 13k + 6 


14 


) 3r 2 + 16r + 21 


15) 3x 2 -17z + 20 


16 


) 3u 2 + 13uv - 10v 2 


17) 3x 2 + 17xy + 102/ 2 


18 


) 7x 2 — 2xy — 5y 2 


19) 5x 2 + 28xy-49y 2 


20 


) 5u 2 + 31uv - 28v 2 


21) 6x 2 -39x-21 


22 


) 10a 2 -54a -36 


23) 21k 2 - 87^-90 


24 


) 21n 2 + 45n-54 


25) 14s 2 -60x + 16 


26 


) 4r 2 + r-3 


27) 6x 2 + 29x + 20 


28 


) 6p 2 +llp-7 


29) 4k 2 -17k + 4 


30 


) 4r 2 + 3r-7 


31) 4x 2 + 9a;y + 2t/ 2 


32 


4m 2 + 6mn + 6n 2 


33) 4m 2 — 9mn — 9n 2 


34 


) Ax 2 -6xy + 30y 2 


35) 4x 2 + 13xy + 3y 2 


36 


) 18u 2 - 3uv - 36v 2 


37) 12x 2 + 62x2/ + 70y 2 


38 


) 16x 2 + 60a;y + 36y 2 


39) 2Ax 2 -52xy + 8y 2 


40 


) 12x 2 + 50a;y + 28?/ 2 



228 



6.5 

Factoring - Factoring Special Products 



Objective: Identify and factor special products including a difference of 
squares, perfect squares, and sum and difference of cubes. 

When factoring there are a few special products that, if we can recognize them, 
can help us factor polynomials. The first is one we have seen before. When multi- 
plying special products we found that a sum and a difference could multiply to a 
difference of squares. Here we will use this special product to help us factor 

Difference of Squares: a 2 — b 2 = (a + b)(a — b) 

If we are subtracting two perfect squares then it will always factor to the sum and 
difference of the square roots. 



Example 300. 

x 2 — 16 Subtracting two perfect squares, the square roots are x and 4 

(x + 4) (x — 4) Our Solution 

Example 301. 

9a 2 — 256 2 Subtracting two perfect squares, the square roots are 3a and 56 
(3a + 56) (3a — 56) Our Solution 



It is important to note, that a sum of squares will never factor. It is always 
prime. This can be seen if we try to use the ac method to factor x 2 + 36. 



Example 302. 

x 2 + 36 No bx term, we use Ox. 

x 2 + 0x + 36 Multiply to 36, add to 

1 • 36, 2 • 18,3 • 12,4- 9, 6 • 6 No combinations that multiply to 36 add to 

Prime, cannot factor Our Solution 



229 



It turns out that a sum of squares is always prime. 

Sum of Squares: a 2 + b 2 = Prime 

A great example where we see a sum of squares comes from factoring a difference 
of 4th powers. Because the square root of a fourth power is a square (va 4 = a 2 ), 
we can factor a difference of fourth powers just like we factor a difference of 
squares, to a sum and difference of the square roots. This will give us two factors, 
one which will be a prime sum of squares, and a second which will be a difference 
of squares which we can factor again. This is shown in the following examples. 

Example 303. 

a 4 — b A Difference of squares with roots a 2 and b 2 
(a 2 + b 2 ) (a 2 — b 2 ) The first factor is prime, the second is a difference of squares! 
(a 2 + b 2 )(a + b)(a-b) Our Solution 

Example 304. 

x 4 — 16 Difference of squares with roots x 2 and 4 
(x 2 + 4) (x 2 — 4) The first factor is prime, the second is a difference of squares! 

(x 2 + 4)(x + 2)(x-2) Our Solution 

Another factoring shortcut is the perfect square. We had a shortcut for multi- 
plying a perfect square which can be reversed to help us factor a perfect square 

Perfect Square: a 2 + 2ab + b 2 = (a + b) 2 

A perfect square can be difficult to recognize at first glance, but if we use the ac 
method and get two of the same numbers we know we have a perfect square. 
Then we can just factor using the square roots of the first and last terms and the 
sign from the middle. This is shown in the following examples. 

Example 305. 

x 2 - 6x + 9 Multiply to 9 , add to - 6 

The numbers are — 3 and — 3, the same! Perfect square 
(x — 3) 2 Use square roots from first and last terms and sign from the middle 

230 



Example 306. 

4x 2 + 20xy + 25y 2 Multiply to 100, add to 20 

The numbers are 10 and 10, the same! Perfect square 
(2x + 5y) 2 Use square roots from first and last terms and sign from the middle 

World View Note: The first known record of work with polynomials comes 
from the Chinese around 200 BC. Problems would be written as "three sheafs of a 
good crop, two sheafs of a mediocre crop, and one sheaf of a bad crop sold for 29 
dou. This would be the polynomial (trinomial) 3x + 2y + z = 29. 

Another factoring shortcut has cubes. With cubes we can either do a sum or a 
difference of cubes. Both sum and difference of cubes have very similar factoring 
formulas 

Sum of Cubes: a 3 + b 3 = (a + b) (a 2 - ab + b 2 ) 

Difference of Cubes: a 3 — b 3 = (a — b) (a 2 + ab + b 2 ) 

Comparing the formulas you may notice that the only difference is the signs in 
between the terms. One way to keep these two formulas straight is to think of 
SOAP. S stands for Same sign as the problem. If we have a sum of cubes, we add 
first, a difference of cubes we subtract first. O stands for Opposite sign. If we 
have a sum, then subtraction is the second sign, a difference would have addition 
for the second sign. Finally, AP stands for Always Positive. Both formulas end 
with addition. The following examples show factoring with cubes. 



Example 


307. 




m - 


-27 




(m 


3)(rr, 


i 2 3m 


9) 




(m — 


3)(m 2 


! + 3m + 9) 



We have cube roots m and 3 

Use formula, use SOAP to fill in signs 

Our Solution 



Example 308. 

125p 3 + 8r 3 We have cube roots 5p and 2r 
(5p 2r)(25p 2 lOr 4r 2 ) Use formula, use SOAP to fill in signs 
(5p + 2r) (25p 2 - lOr + 4r 2 ) Our Solution 

The previous example illustrates an important point. When we fill in the trino- 
mial's first and last terms we square the cube roots 5p and 2r. Often students 
forget to square the number in addition to the variable. Notice that when done 
correctly, both get cubed. 

231 



Often after factoring a sum or difference of cubes, students want to factor the 
second factor, the trinomial further. As a general rule, this factor will always be 
prime (unless there is a GCF which should have been factored out before using 
cubes rule). 

The following table sumarizes all of the shortcuts that we can use to factor special 
products 

Factoring Special Products 



Difference of Squares 


a 2 -b 2 =(a + b)(a-b) 


Sum of Squares 


a 2 + b 2 = Prime 


Perfect Square 


a 2 + 2ab + b 2 = (a + b) 2 


Sum of Cubes 


a 3 + b 3 = (o + b)(a 2 - ab + b 2 


Difference of Cubes 


a 3 - b 3 = (a - b)(a 2 + ab + b 2 



As always, when factoring special products it is important to check for a GCF 
first. Only after checking for a GCF should we be using the special products. 
This is shown in the following examples 

Example 309. 

72a: 2 -2 GCF is 2 
2(36x 2 — 1) Difference of Squares, square roots are 6x and 1 
2(6x + 1) (6a; — 1) Our Solution 

Example 310. 



48a; 2 y - 24a;y - 
3y(16x 2 -8x 



3y GCF is 3?/ 

-1) Multiply to 16 add to 8 

The numbers are 4 and 4, the same! Perfect Square 
3y(4x — l) 2 Our Solution 



Example 311. 



128a 4 6 2 + 54a6 5 

2afe 2 (64a 3 + 276 3 ) 

2ab 2 (4a + 36) (16a 2 - Ylab + 9b 2 ) 



GCFis2a6 2 

Sum of cubes! Cube roots are 4a and 3b 

Our Solution 



232 



6.5 Practice - Factoring Special Products 



Factor each completely. 

1) r 2 -16 
3) t> 2 -25 
5) p 2 - 4 
7) 9k 2 - 4 
9) Sx 2 - 27 



11 
13 

15 
17 
19 
21 
23 
25 
27 
29 
31 
33 
35 
37 
39 
41 
43 
45 
47 



16x 2 -36 
18a 2 - 506 2 
a 2 - 2a + 1 
x 2 + 6x + 9 
x 2 — 6x + 9 
25p 2 -10p + l 
25a 2 + 30a6 + 96 2 
4a 2 -20afe + 25fc 2 
8x 2 -24xy+18y 2 
8 — m 3 
x 3 - 64 
216 -u 3 
125a 3 -64 
64a: 3 + 27 y 3 
54x 3 + 250y 3 
a 4 -81 
16 -z 4 



4 4 

x — y 



2) 


x 2 -9 


4) 


x 2 -l 


6) 


4w 2 -l 


8) 


9a 2 -1 


10 


) 5n 2 - 20 


12 


) 125x 2 + 45t/ 2 


14 


) 4m 2 + 64n 2 


16 


) k 2 + 4k + 4 


18 


) n 2 — 8n + 16 


20 


) k 2 - 4k + 4 


22 


) x 2 + 2x + 1 


24 


) x 2 + 8xy + 16y 2 


26 


) 18m 2 — 24mn + 8n 2 


28 


) 20x 2 + 20xy + 5y 2 


30 


) x 3 + 64 


32 


) x 3 + 8 


34 


) 125x 3 -216 


36 


) 64x 3 - 27 


38 


) 32m 3 - 108n 3 


40 


) 375m 3 + 648n 3 


42 


) x 4 -256 


44 


) n 4 -l 


46 


) 16a 4 -b 4 


48 


) 81c 4 -16d 4 



m 4 - 816 4 



233 



6.6 



Factoring - Factoring Strategy 



Objective: Identity and use the correct method to factor various poly- 
nomials. 

With so many different tools used to factor, it is easy to get lost as to which tool 
to use when. Here we will attempt to organize all the different factoring types we 
have seen. A large part of deciding how to solve a problem is based on how many 
terms are in the problem. For all problem types we will always try to factor out 
the GCF first. 

Factoring Strategy (GCF First!!!!!) 

• 2 terms: sum or difference of squares or cubes: 
a 2 -b 2 = (a + b)(a-b) 

a 2 + b 2 = Prime 

a 3 + b 3 = (a + b) (a 2 - ab + b 2 ) 

a 3 - 6 3 = (a - b) (a 2 + ab + b 2 ) 

• 3 terms: ac method, watch for perfect square! 
a 2 + 2ab + b 2 = (a + b) 2 

Multiply to ac and add to b 

• 4 terms: grouping 

We will use the above strategy to factor each of the following examples. Here the 
emphasis will be on which strategy to use rather than the steps used in that 
method. 

Example 312. 

4x 2 + 56xy + l96y 2 GCF first, 4 

4(x 2 + 14xy + 49y 2 ) Three terms, try ac method, multiply to 49, add to 14 

7 and 7, perfect square! 



234 



4(x + 7y) 2 Our Solution 



Example 313. 



5x 2 y + 15xy — 35x 2 — 105x GCF first, 5x 

5x(xy + Sy — 7x — 21) Four terms, try grouping 

5x[y(x + 3) — 7(x + 3)] (x + 3) match! 

5x(x + 3)(y — 7) Our Solution 



Example 314. 

100x 2 -400 GCF first, 100 
100(x 2 — 4) Two terms, difference of squares 
100(x + 4) (x - 4) Our Solution 

Example 315. 

108a; 3 |/ 2 -39xV + 3a;2/ 2 GCF first, 3a; y 2 

3xy 2 (36x 2 — 13x + 1) Thee terms, ac method, multiply to 36, add to — 13 

3xy 2 (36x 2 — 9x — 4x + 1) — 9 and — 4, split middle term 

3xy 2 [9x(4x — 1) — l(4a; — 1)] Factor by grouping 

3xy 2 (4x — l)(9x — 1) Our Solution 

World View Note: Variables originated in ancient Greece where Aristotle would 
use a single capital letter to represent a number. 

Example 316. 



5 + 625y 3 

5(l + 125y 3 ) 
5(l + 5y)(l-5y + 25y 2 ) 



GCF first, 5 

Two terms, sum of cubes 

Our Solution 



It is important to be comfortable and confident not just with using all the fac- 
toring methods, but decided on which method to use. This is why practice is very 
important! 



235 



6.6 Practice - Factoring Strategy 



Factor 


each completely. 


1) 


24a z 


-18ah + 60yz-45yh 


3) 


5u 2 - 


9uv + 4i> 2 


5) 


-2x 


3 + 128y 3 


7) 


5n 3 + In 2 — 6n 


9) 


54u 3 


-16 


11 


) n 2 - 


- n 


13 


)x 2 - 


4xy + 3y 2 


15 


) 9x 2 


-25y 2 


17 


2 

m 


-An 2 


19 


) 36b 2 


c-16xd-24b 2 d + 24:xc 


21 


) 128 + 54a; 3 


23 


) 2x 3 


-f 6x 2 y — 20y 2 a; 


25 


) n 3 + 7n 2 + 10n 


27 


) 21x 


3 -64 


29 


) 5x 2 


+ 2x 


31 


) 3k 3 


-27fc 2 + 60fc 


33 


ran 


— 12a; + 3m — 4a;n 


35 


16a: 


2 - 8xy + y 2 


37 


) 27rr 


2 -A8n 2 


39 


) 9x 3 


-\-21x 2 y — 60y 2 x 


41 


) 2m 2 


+ 6mn — 20n 2 



2) 


2x 2 -llx + 15 


4) 


16x 2 + 48xy + 36y 2 


6) 


20uv — 60m 3 — 5xv + 15xu 2 


8) 


2x 3 + 5x 2 y + 3y 2 x 


10 


) 54 -128a; 3 


12 


) 5a; 2 -22a; -15 


14 


) 45u 2 -150uv + 125v 2 


16 


) x 3 -27y 3 


18 


) 12ab-18a + 6nb-9n 


20 


3m 3 — 6m 2 n — 24n 2 m 


22 


) 64m 3 + 27n 3 


24 


) 3ac + 15a d 2 + x 2 c + 5a; 2 d 2 


26 


64m 3 — n 3 


28 


) 16a 2 - 96 2 


30 


) 2x 2 -10a; + 12 


32 


) 32x 2 -18y 2 


34 


)2k 2 + k-10 


36 


) v 2 + v 


38 


) a; 3 + 4x 2 


40 


) 9n 3 - 3n 2 


42 


) 2u 2 v 2 -lluv 3 + l5v 4 



236 



6.7 

Factoring - Solve by Factoring 



Objective: Solve quadratic equation by factoring and using the zero 
product rule. 

When solving linear equations such as 2x — 5 = 21 we can solve for the variable 
directly by adding 5 and dividing by 2 to get 13. However, when we have x 2 (or a 
higher power of x) we cannot just isolate the variable as we did with the linear 
equations. One method that we can use to solve for the varaible is known as the 
zero product rule 

Zero Product Rule: If a b = then either a = or b = 

The zero product rule tells us that if two factors are multiplied together and the 
answer is zero, then one of the factors must be zero. We can use this to help us 
solve factored polynomials as in the following example. 



Example 317. 



(2x — 3) (5x + 1) = One factor must be zero 

2x — 3 = or 5a; + 1 = Set each factor equal to zero 

+ 3 + 3 —1 — 1 Solve each equation 

2x = 3 or 5x = — 1 



"2" T~ ~5~ "5 
3 

2 



3 — 1 

x = — or — — Our Solution 



For the zero product rule to work we must have factors to set equal to zero. This 
means if the problem is not already factored we will factor it first. 



Example 318. 

4x 2 + x — 3 = Factor using the ac method, multiply to — 12, add to 1 

4x 2 — 3x + 4x — 3 = The numbers are — 3 and 4, split the middle term 

x(4x — 3) + l(4x — 3) = Factor by grouping 

(4x — 3)(x+ 1) = One factor must be zero 

4x — 3 = or £ + 1 = Set each factor equal to zero 



237 



3 + 3 



Ax = 3 or x 

~4~ ~4~ 

3 



4 



Solve each equation 



or — 1 Our Solution 



Another important part of the zero product rule is that before we factor, the 
equation must equal zero. If it does not, we must move terms around so it does 
equal zero. Generally we like the x 2 term to be positive. 



Example 319. 






x 2 = 


--8x- 


15 


-8x + 15 - 


-8^ + 15 


x 2 — 8x 


+ 15 = 


= 


(x — 5)(x 


-3) = 


= 


x — 5 = or : 


r — 3 - 


= 


+ 5 + 5 


+ 3 + 3 



5 or x = 3 



Set equal to zero by moving terms to the left 

Factor using the ac method , multiply to 1 5 , add to 
The numbers are — 5 and — 3 
Set each factor equal to zero 
Solve each equation 
Our Solution 



Example 


320. 






(x — 


7)(z + 3) = 


-9 


x 2 — 7x + 3x - 


-21 = 


-9 


x 2 - 


-4x- 


-21 = 


-9 






+ 9 


+ 9 


2 


; 2 -4x-12: 


= 


(x 


-6)0 


r + 2) 


= 


x — 6 = 


-0 or 


x + 2 


= 


+ 6 + 6 


-2 


-2 




x = 


6 or 


-2 


Example 


321. 







Not equal to zero, multiply first, use FOIL 

Combine like terms 

Move — 9 to other side so equation equals zero 

Factor using the ac method, mutiply to — 12, add to — 4 
The numbers are 6 and — 2 
Set each factor equal to zero 
Solve each equation 
Our Solution 



3x 2 + 4x — 5 = 7x 2 + 4x — 14 Set equal to zero by moving terms to the right 
Sx 2 - Ax + 5 - Sx 2 - 4x + 5 



= 4x 2 — 9 Factor using difference of squares 



238 



= (22 + 3)(22-3) 

2a; + 3 = or 22-3 = 

-3-3 +3+3 



2x~- 


= — 3 or 2a: = 3 


~T 


~T ~T ~2~ 




-3 3 

x = —— or — 

2 2 



One factor must be zero 
Set each factor equal to zero 
Solve each equation 



Our Solution 



Most problems with x 2 will have two unique solutions. However, it is possible to 
have only one solution as the next example illustrates. 



Example 


322. 








4a: 2 


= 12a:- 


-9 


-122 + 9 


-122 + 9 


4a; 2 - 12a; + 9 : 


= 




(2x 


-3) 2 : 


= 






t 


2x — 3 = 


= 






+ 3 + 3 






22 = 


= 3 






~T 


~T 






x = 


3 

= 2 



Set equal to zero by moving terms to left 

Factor using the ac method, multiply to 36, add to — 12 
— 6 and — 6, a perfect square! 
Set this factor equal to zero 
Solve the equation 



Our Solution 



As always it will be important to factor out the GCF first if we have one. This 
GCF is also a factor and must also be set equal to zero using the zero product 
rule. This may give us more than just two solution. The next few examples illus- 
trate this. 



Example 


323. 










4x 2 = 


82 






-82- 


82 




42 2 


-82: 


= 




42(a 


:-2) = 


= 


42 = 


= or 


2-2: 


= 


~4~ 


~4~ 


+ 2 + 2 




a; 


= or 2 



Set equal to zero by moving the terms to left 

Be careful, on the right side, they are not like terms! 

Factor out the GCF of 42 

One factor must be zero 

Set each factor equal to zero 

Solve each equation 

Our Solution 



239 



Example 324. 





2x 3 - Ux 2 + 24x = 


= 




2x(x 2 - 


-7a; + 12) = 


= 




2x(x — 


3) (a: 


-4) = 


= 


2x -- 


= or x — 3 = 


or i 


:-4 = 


= 


~2 


~2~ +3 + 3 




+ 4 + 4 




x = or 


3 or 


4 




Example 325. 










6x 2 + 21x - 


-27 = 


= 




3(2x 2 


+ 7x- 


-9) = 


= 




3(2x 2 + 9x 


-2x- 


-9) = 


= 




3[x(2x + 9)-: 


l(2x + 9)] = 


= 




3(2x + 


9)(x- 


-1) = 


= 


3 = 


or 2x + 9 = 


or x 


-1 = 


= 


H 


-9-9 




+ 1 + 1 




2x = 


— 9 or x = 


= 1 




~2 


X 







Factor out the GCF of 2x 

Factor with ac method, multiply to 12, add to — 7 

The numbers are — 3 and — 4 

Set each factor equal to zero 

Solve each equation 

Our Solutions 



Factor out the GCF of 3 

Factor with ac method, multiply to 

The numbers are 9 and — 2 

Factor by grouping 

One factor must be zero 

Set each factor equal to zero 

Solve each equation 



18, add to 7 



x - 



or 1 Our Solution 



In the previous example, the GCF did not have a variable in it. When we set this 
factor equal to zero we got a false statement. No solutions come from this factor. 
Often a student will skip setting the GCF factor equal to zero if there is no vari- 
ables in the GCF. 

Just as not all polynomials cannot factor, all equations cannot be solved by fac- 
toring. If an equation does not factor we will have to solve it using another 
method. These other methods are saved for another section. 

World View Note: While factoring works great to solve problems with x 2 , 
Tartaglia, in 16th century Italy, developed a method to solve problems with x 3 . 
He kept his method a secret until another mathematician, Cardan, talked him out 
of his secret and published the results. To this day the formula is known as 
Cardan's Formula. 

A question often asked is if it is possible to get rid of the square on the variable 
by taking the square root of both sides. While it is possible, there are a few prop- 
erties of square roots that we have not covered yet and thus it is common to 
break a rule of roots that we are not aware of at this point. The short reason we 
want to avoid this for now is because taking a square root will only give us one of 
the two answers. When we talk about roots we will come back to problems like 
these and see how we can solve using square roots in a method called completing 
the square. For now, never take the square root of both sides! 



240 



6.7 Practice - Solve by Factoring 



Solve each equation by factoring. 

1) (Jfe-7)(Jfe + 2) = 
3) (x-l)(x + 4) = 

5) 6x 2 - 150 = 

7) 2n 2 + 10n-28 = 

9) 7x 2 + 26^ + 15 = 



11 


) 5n 2 - 9n - 2 = 


13 


) x 2 - Ax - 8 = - 8 


15 


x 2 — 5x — 1 = — 5 


17 


) 49p 2 + 371p-163 = 5 


19 


) 7x 2 + 17x - 20 = - 8 


21 


) 7r 2 + 84 = - 49r 


23 


) x 2 — 6x = 16 


25 


) 3 V 2 + 7 V = 40 


27 


) 35x 2 + 120x = -45 


29; 


4/e 2 +18/c-23 = 6/c-7 


3i; 


9x 2 - 46 + 7x = 7x + 8x 2 + 


33; 


2m 2 + 19m + 40 = - 2m 


35; 


40p 2 + 183p - 168 = p + 5p 



2) (a + 4)(o-3) = 

4) (2x + 5)(x-7) = 
6) p 2 + Ap - 32 = 
8) m 2 - m - 30 = 



10 


) 40r 2 -285r- 280 = 


12 


) 26 2 - 36 - 2 = 


14 


) v 2 -8w-3 = -3 


16 


) a 2 -6a + 6 = -2 


18 


) 7A; 2 + 57A; + 13 = 5 


20 


) 4n 2 -13n + 8 = 5 


22 


) 7m 2 - 224 = 28m 


24 


) 7n 2 -28n = 


26 


) 6b 2 = 5 + lb 


28 


) 9n 2 + 39n = -36 


30 


) a 2 + 7a - 9 = - 3 + 6a 


32 


) x 2 + 10x + 30 = 6 


34 


) 5n 2 + 41n + 40 = -2 


36 


) 24x 2 + llx-80 = 3x 



241 



Chapter 7 : Rational Expressions 

7.1 Reduce Rational Expressions 243 

7.2 Multiply and Divide 248 

7.3 Least Common Denominator 253 

7.4 Add and Subtract 257 

7.5 Complex Fractions 262 

7.6 Proportions 268 

7.7 Solving Rational Equations 274 

7.8 Application: Dimensional Analysis 279 



242 



7.1 



Rational Expressions - Reduce Rational Expressions 



Objective: Reduce rational expressions by dividing out common fac- 
tors. 

Rational expressions are expressions written as a quotient of polynomials. 
Examples of rational expressions include: 



x 2 — x — 12 . 3 , a — b .3 

— = — — and and and — 

s 2 -9a; + 20 x-2 b-a 2 



As rational expressions are a special type of fraction, it is important to remember 
with fractions we cannot have zero in the denominator of a fraction. For this 
reason, rational expressions may have one more excluded values, or values that 
the variable cannot be or the expression would be undefined. 



Example 326. 



State the excluded value(s) : 



1 



3x 2 + 5x 



Denominator cannot be zero 



Sx 2 + 5x^0 Factor 

x(3x + 5) ^ Set each factor not equal to zero 
x=^0 or 3x + 5^0 Subtract 5 from second equation 
-5-5 

Divide by 3 



3x^ — 5 



i^O or 



5 



Second equation is solved 
Our Solution 



This means we can use any value for x in the equation except for and — . We 



243 



can however, evaluate any other value in the expression. 

World View Note: The number zero was not widely accepted in mathematical 
thought around the world for many years. It was the Mayans of Central America 
who first used zero to aid in the use of their base 20 system as a place holder! 

Rational expressions are easily evaluated by simply substituting the value for the 
variable and using order of operations. 



Example 327. 



x 2 -4 



x 2 + 6x + 8 



when x = — 6 Substitute — 5 in for each variable 



(-6) 2 -4 



(-6) 2 + 6(-6) + 8 

36-4 

36 + 6(-6) + 8 

36-4 

36-36 + 8 

32 



Exponents first 

Multiply 

Add and subtract 

Reduce 



4 Our Solution 



Just as we reduced the previous example, often a rational expression can be 
reduced, even without knowing the value of the variable. When we reduce we 
divide out common factors. We have already seen this with monomials when we 
discussed properties of exponents. If the problem only has monomials we can 
reduce the coefficients, and subtract exponents on the variables. 



Example 328. 



15x v 

tt— 77 Reduce , subtract exponents. Negative exponents move to denominator 

25x^y b 

3x 2 



5y 4 



Our Solution 



244 



However, if there is more than just one term in either the numerator or denomi- 
nator, we can't divide out common factors unless we first factor the numerator 
and denominator. 



Example 329. 



28 



8x 2 - 


16 


28 




8(x 2 - 


■2) 


7 





2(x 2 -2) 



Denominator has a common factor of I 



Reduce by dividing 24 and 8 by 4 



Our Solution 



Example 330. 



9a: -3 

18a; -6 



Numerator has a common factor of 3, denominator of 6 



3(3x-i; 
6(3x-i; 



Divide out common factor (3x — 1) and divide 3 and 6 by 3 



1 

2 



Our Solution 



Example 331. 



a: 2 -25 



x 2 + 8x 


+ 15 


(x + 5)(x 


-5) 


(x + 3)(x 


+ 5) 




E — 5 



a; + 3 



Numerator is difference of squares, denominator is factored using ac 



Divide out common factor (x + 5) 



Our Solution 



It is important to remember we cannot reduce terms, only factors. This means if 
there are any + or — between the parts we want to reduce we cannot. In the pre- 



x — 5 



vious example we had the solution — -, we cannot divide out the x's because 



x + 3 



they are terms (separated by + or — ) not factors (separated by multiplication). 



245 



7.1 Practice - Reduce Rational Expressions 



Evaluate 

1) -^- — when v = 4 

a; — 3 



3) 
5) 



c 2 - 4a; + 3 
6 + 2 



when a; = — 4 
when 6 = 



(,2 + 4ft + 4 

State the excluded values for each. 



7) 
9) 
11 



3fc 2 + 30fc 
k + 10 

15n 2 
10n + 25 

, 10m 2 + 8m 
' 10m 



13) 
15) 



r 2 + 3r + 2 
5r + 10 

6 2 + 126 + 32 



6 2 + 46 - 32 

Simplify each expression. 

17) 21x2 

19) 

21 

23) 

25) 

27) 

29) 

31 

33) 

35) 

37) 

39 ) <)„-uo 



18a; 


24a 


40a 2 


32a- 3 


8a- 4 


18m - 24 


60 


20 


4p + 2 


2+1 


a; 2 + 8a; + 7 


32a; 2 


28a- 2 + 28a; 


n 2 + 4n- 12 


n 2 - 7n + 10 


9i> + 54 


v 2 - 4t) - 60 


12a- 2 - 42a; 


30a- 2 - 42a; 


6a- 10 


10a + 4 


2n 2 + 19n - 10 



2) 


b- 

36 


-9 Wli 


4) 




a + 2 


a 2 


+ 3a + 2 


k\ 


n 2 


— n — 6 



n — 3 



27p 



m 6 = - 2 
when a = — 1 
when n = 4 



10 
12 
14 
16 

18 
20 
22 
24 
26 
28 
30 
32 
34 
36 
38 
40 



18p 2 - 36p 


, 2+10 


' 8a; 2 + 80a; 


^ 10a; +16 


' 6a; + 20 


^ 6n 2 - 21n 


6n 2 + 3n 


^ 10u 2 + 30u 



35« 2 — 5v 



12n 

4n 2 



21k 


24k 2 


90a; 2 


20a- 


10 


81n 3 + 36n 2 


n-9 


9n-81 


28m + 12 


36 


49r + 56 


56r 


b 2 + 14b + 48 


b 2 + 156 + 56 


30a; - 90 


50a; + 40 


k 2 - 12k + 32 


fc 2 -64 


9p + 18 


p 2 + 4p + 4 


3a- 2 - 29a; + 40 



5a; 2 - 30a; - 80 



246 



41) 
43) 

45) 
47) 
49) 



8m + 16 


20m 


-12 


2x 2 - 


- ICte + 8 


3x 2 


- 7z + 4 


7n 2 - 


- 32n + 16 


4n-16 


n 2 - 


2n + l 


6n + 6 


7a 2 - 


- 26a - 45 



6a 2 - 34a + 20 



42) 
44) 
46) 
48) 
50) 



56x - 48 



24x 2 + 56x + 32 


506 - 80 


50fe + 20 


35v + 35 


21ij + 7 


56x - 48 


24x 2 + 56x + 32 


4fc 3 - 2fe 2 - 2k 



9k 3 - 18fc 2 + 9k 



247 



7.2 

Rational Expressions - Multiply & Divide 

Objective: Multiply and divide rational expressions. 

Multiplying and dividing rational expressions is very similar to the process we use 
to multiply and divide fractions. 

Example 332. 

— - • -— First reduce common factors from numerator and denominator (15 and 7) 
49 45 v ' 



1 2 

7'3 

2 
21 



Multiply numerators across and denominators across 



Our Solution 



The process is identical for division with the extra first step of multiplying by the 
reciprocal. When multiplying with rational expressions we follow the same pro- 
cess, first divide out common factors, then multiply straight across. 



248 



Example 333. 

25x 2 24y 4 Reduce coefficients by dividing out common factors (3 and 5) 
9y 8 55x 7 Reduce, subtracting exponents, negative exponents in denominator 

5 8 



3y 4 llx 5 

40 



33x 5 y 4 



Multiply across 



Our Solution 



Division is identical in process with the extra first step of multiplying by the 
reciprocal. 



Example 334. 

a% 2 b 4 



a ' 4 

a% 2 \_ 
a ' 6 4 

a 3 4 
T'fe2 

4a 3 



Multiply by the reciprocal 

Subtract exponents on variables, negative exponents in denominator 

Multiply across 



Our Solution 



Just as with reducing rational expressions, before we reduce a multiplication 
problem, it must be factored first. 



Example 335. 

x 2 -9 x 2 -8x + 16 _ , ,, 

b actor each numerator and denominator 



Divide out common factors (x + 3) and (x — 4) 
Multiply across 





X 2 + X 


-20 


3x + 9 




(x + S)(x — 


3) 


(x — 4)(x — 


4) 


(x - 


-4)(z + 


5) 


3(a; + 3) 

Jb O Jb ~ 


-4 



x + 5 3 



249 



(x-3)(x-4) 
3(x + 5) 



Our Solution 



Again we follow the same pattern with division with the extra first step of multi- 
plying by the reciprocal. 



Example 336. 




x 2 -x-12 


5x 2 + 15x 


x 2 -2x-8 ' 


x 2 + x — 2 


x 2 -x-12 


x 2 + x — 2 



x z 



2x — 8 5x 2 +15x 



(x-4)(x + 3) (x + 2)(x-l) 
(x + 2)(x-4)' 5x(x + 3) 

1 x-1 

1 5x 

x — 1 



5.x 



Multiply by the reciprocal 

Factor each numerator and denominator 

Divide out common factors: 
(x — 4) and (x + 3) and (x + 2) 
Multiply across 

Our Solution 



We can combine multiplying and dividing of fractions into one problem as shown 
below. To solve we still need to factor, and we use the reciprocal of the divided 
fraction. 



Example 337. 

a 2 + la + 10 



o+l 



a — 1 



a" 



6a + 5 a 2 + 4a + 4 



(a + 5)(a + 2) (o + l) . (a-1) 

(o + 5)(o+l) '(a + 2)(a+2) ' (a + 2) 

(a + 5)(a + 2) (o + l) (a + 2) 

(o + 5)(o+l) ' (a + 2)(a + 2) ' (a-1) 



a-1 



Factor each expression 
Reciprocal of last fraction 

Divide out common factors 

(a + 2), (a + 2), (o+l), (a + 5) 
Our Solution 



World View Note: Indian mathematician Aryabhata, in the 6th century, pub- 
lished a work which included the rational expression — — ^— — - for the sum of 
the first n squares (1 1 + 2 2 + 3 2 + .... + n 2 ) 



250 



7.2 Practice - Multiply and Divide 



Simplify each expression. 

1) 

9n 7 



8x 2 9 
~9~ ' 2 



3) 
5) 
7) 
9) 
11 
13 
15 
17 
19 
21 
23 
25 
27 
29 
31 
33 
35 
37 
39 



2n 5n 

5x 2 6 
~T~ ' 5 

7(m — 6) 5m(7m-5) 



m — 6 

7r 



7(7m - 5) 
r — 6 



7r(r+10) ' (r-6) 2 

25n + 25 4 

5 30n + 30 

x-10 7 



35x + 21 " 35x + 21 
a; 2 — 6x — 7 x + 5 



i + 5 



x-7 



24k 2 -40k " 15fc-25 
6 



(n 



' lOn - 80 
4m + 36 m — 5 



m + 9 

3x — 6 
12a; - 24 

6 + 2 



5m 2 

(x + 3) 
(56-3) 



406 2 - 24fc v 
n - 1 12 - 6n 



6n - 12 n 2 - 13n + 42 




27a + 36 . 6a + 8 




9a + 63 ' 2 




a; 2 - 12a; + 32 7a: 2 + 14a; 




a; 2 - 6a; - 16 7a; 2 + 21a; 




(10m 2 + 100m) .^i 


-36m 2 
-40m 


7p 2 + 25p +12 3p - 8 





6p + 48 21p 2 - 44p - 32 

10b 2 30b + 20 



30b + 20 2b 2 + 10b 
7r 2 -53r-24 . 49r + 21 



7r + 2 



49r + 14 



4) 
6) 
8) 
10 
12 
14 
16 
18 
20 
22 
24 
26 
28 
30 
32 
34 
36 

38 
40 



8a- . 4 

"3lr~ 7 

9m 7 

5m 2 ' 2 



lOp 



7 


n-1 


10(n + 3) ' 


(n + 3)(n-2) 


6x(x + 4) 


(x-3)(x-6) 


x — 3 


6x(x — 6) 


9 


6-5 



6 2 - 6 - 12 ' 6 2 - 6 - 12 
u-1 4 



4 t; 2 - Hi; + 10 
1 8a + 80 



a — 6 

p-8 



p 2 - 12p + 32 " p - 10 
a; 2 - 7a; + 10 x + 10 



x-2 



■a; -20 



2r 



2r 



r + 6 7r + 42 

2n 2 - 12n - 54 . 

ri + 7 ' 

21v 2 + 16v - 16 . 35f - 20 



(2n + 6) 



3v + 4 ' -u - 9 




a; 2 + 11a; + 24 6x 3 + 6x 2 




6x 3 + 18x 2 x 2 + 5a; - 24 




k - 7 7k 2 - 28k 




k 2 - k - 12 8fc 2 - 56fc 




9x 3 + 54x 2 x 2 + 5x - 14 




x 2 + 5x - 14 10x 2 




n-7 . 9n + 54 




n 2 -2n-35 ' 10n + 50 




7x 2 - 66x + 80 . 7x 2 + 39x - 


70 


49x 2 + 7x-72 • 49x 2 + 7x . 


72 


35n 2 - 12n - 32 7n 2 + 16n - 


15 


49n 2 - 91ra + 40 5n + 4 




12x + 24 15x + 21 





10x 2 + 34x + 28 



251 



41] 



x z -l 
2x-A 



3x — 6 



42) 



a s + b s 3a -6b . a 2 - 4fc 2 

-3ab + 26 2 ' 3a 2 -3ab + 3b 2 ~ a + 2b 



43) 



a: 2 + 3x + 9 x 2 + 2x - 8 
x 2 + x-12 ' z 3 -27 



c 2 -6x + 9 



44) 



a: 2 + 3a: - 10 2a; 2 - x - 3 



_ 8x + 20 
x 2 + 6a; + 5 ' 2x 2 + x - 6 ' 6a- +15 



252 



7.3 

Rational Expressions - Least Common Denominators 

Objective: Identity the least common denominator and build up 
denominators to match this common denominator. 

As with fractions, the least common denominator or LCD is very important to 
working with rational expressions. The process we use to find the LCD is based 
on the process used to find the LCD of intergers. 

Example 338. 

Find the LCD of 8 and 6 Consider multiples of the larger number 

8,16, 24. ... 24 is the first multiple of 8 that is also divisible by 6 
24 Our Solution 

When finding the LCD of several monomials we first find the LCD of the coeffi- 
cients, then use all variables and attach the highest exponent on each variable. 

Example 339. 

Find the LCD of 4x 2 y 5 and 6x 4 y 3 z 6 

First find the LCD of coefficients 4 and 6 
12 12 is the LCD of 4 and 6 
x A y 5 z 6 Use all variables with highest exponents on each variable 
12x A y 5 z 6 Our Solution 

The same pattern can be used on polynomials that have more than one term. 
However, we must first factor each polynomial so we can identify all the factors to 
be used (attaching highest exponent if necessary). 

Example 340. 

Find the LCD of x 2 + 2x — 3 and x 2 — x — 12 Factor each polynomial 

(x — 1) (x + 3) and (x — 4) (x + 3) LCD uses all unique factors 
(x — l)(x + 3)(x — 4) Our Solution 

Notice we only used (x + 3) once in our LCD. This is because it only appears as a 
factor once in either polynomial. The only time we need to repeat a factor or use 
an exponent on a factor is if there are exponents when one of the polynomials is 
factored 



253 



Example 341. 

Find the LCD of x 2 - ICte + 25 and x 2 - Ux + 45 



(x — 5) 2 and (x — 5)(x — 9) 
(x-5) 2 (x-9) 



Factor each polynomial 

LCD uses all unique factors with highest exponent 

Our Solution 



The previous example could have also been done with factoring the first polyno- 
mial to (x — 5)(x — 5). Then we would have used (x — 5) twice in the LCD 
because it showed up twice in one of the polynomials. However, it is the author's 
suggestion to use the exponents in factored form so as to use the same pattern 
(highest exponent) as used with monomials. 

Once we know the LCD, our goal will be to build up fractions so they have 
matching denominators. In this lesson we will not be adding and subtracting frac- 
tions, just building them up to a common denominator. We can build up a frac- 
tion's denominator by multipliplying the numerator and denoinator by any factors 
that are not already in the denominator. 



Example 342. 

5 a 



3a 2 b 6a 5 6 3 



Identity what factors we need to match denominators 



la b 3-2 = 6 and we need three more a's and two more b's 

- — I - oin ) Multiply numerator and denominator by this 
3a 2 b\2a 3 b 2 J yy y 



10a 4 fr 2 

6a 5 6 3 



Our Solution 



Example 343. 

x-2 



x + 4 x 2 + lx + Yl 
(x + 4)(x + 3) 



Factor to identity factors we need to match denominators 



(x + 3) The missing factor 



X 


" 2 ( 


'x 


+ 3\ 


X 


+ 4l 


+ sj 




x 2 + x 


-6 



(x + 4)(x + 3) 



Multiply numerator and denominator by missing factor, 
FOIL numerator 
Our Solution 



254 



As the above example illustrates, we will multiply out our numerators, but keep 
our denominators factored. The reason for this is to add and subtract fractions 
we will want to be able to combine like terms in the numerator, then when we 
reduce at the end we will want our denominators factored. 

Once we know how to find the LCD and how to build up fractions to a desired 
denominator we can combine them together by finding a common denominator 
and building up those fractions. 

Example 344. 

Build up each fraction so they have a common denominator 

and 9 First identify LCD 



46 3 c 6a 2 b 

12a 2 b 3 c Determine what factors each fraction is missing 
First: 3a 2 Second: 2b 2 c Multiply each fraction by missing factors 



_5a_(3a?\ 3c (2b 2 c 

4b 3 c\3a 2 ) 6a 2 b\2b 2 c 



15a 3 . 6b 2 c 2 „ n . J . 

and „ - 010 Our Solution 



12a 2 6 3 c 12a 2 b 3 c 

Example 345. 

Build up each fraction so they have a common denominator 

and — ^ Factor to find LCD 



x 2 — 5x — 6 x 2 + Ax + 3 

(x — 6)(x + 1) (x + l)(x + 3) Use factors to find LCD 

LCD: {x — 6)(x + l)(x + 3) Identify which factors are missing 

First: (x + 3) Second: (x — 6) Multiply fractions by missing factors 



^\ rp j np I X \ rp i if fi \ 

and — -r- Multiply numerators 



(x-6){x + l)\x + 3J {x+l)(x + 3)\x-6 



5x 2 +15x , a; 2 -8x + 12 ^ D , ,. 

and -. — —7 — Our Solution 



(a;-6)(a; + l)(x + 3) (cc-6)(a; + l)(x + 3) 

World View Note: When the Egyptians began working with fractions, they 

expressed all fractions as a sum of unit fraction. Rather than -, they would write 

ill 
the fraction as the sum, - + - + — . An interesting problem with this system is 

this is not a unique solution, - is also equal to the sum ^ + 7^ + 77 + 777. 

1 5 x 3 5 D 10 



255 



7.3 Practice - Least Common Denominator 



Build up 


denominators. 


1) 


3 _ ? 

8 ~~ 48 




3) 


a _ ? 
x xy 




5) 


2 
3a 3 b 2 c 


? 


9a 5 6 2 c 4 


7) 


2 
x + 4 a 


? 


; 2 -16 


Ql 


x — 4 


? 



2) 


a V 

5 5 a 






11 


5 
2a- 2 5 


? 




i 1 


lx 3 y 




6) 


4 




? 


3a 6 ft 2 c 4 


9i 


!56 2 C 4 


8) 


x + l 




? 


x — 3 


a; 2 - 


■6a: + 9 


in 


^ a; — 6 




7 



a; + 2 x 2 + 5x + 6 ' a; + 3 x 2 - 2x - 15 

Find Least Common Denominators 

11) 2a 3 ,6a 4 b 2 ,4a 3 b 5 12) 5x 2 t/, 25x 3 y 5 ^ 

13) x 2 - 3x, x - 3, x 14) 4x - 8, x - 2, 4 

15)x + 2,x-4 16) x,x-7,x + l 

17) x 2 - 25, x + 5 18) x 2 - 9, x 2 - 6x + 9 

19) x 2 + 3x + 2, a; 2 + 5x + 6 20) x 2 - 7x + 10, x 2 - 2a; - 15, x 2 + x - 6 

Find LCD and build up each fraction 

r.r.\ 3x 2 





5b 2 ' 10a 3 b 




23) 


x + 2 x 
x — 3 ' x 


-3 

+ 2 




25) 


x 

a; 2 -16' 




3 a; 


a; 2 - 


-8a; +16 


27) 


x + l 

x 2 - 36 ' 




2a- + 3 


a; 2 - 


f 12x + 36 


29) 


4x 


6' 


r + 2 
r-3 



Z,Z,J 


1-4' x + 2 




24) 


5 2 
a; 2 — 6a; ' a; ' 


-3 

x — 6 


26) 


5a; + 1 
a: 2 - 3a: - 10 


4 
' x-5 


28) 


3a; + 1 
a; 2 -a;- 12' 


2x 


x 2 + 4x + 3 


30) 


3a p 


x-2 

„2 i „ on 



256 



7.4 

Rational Expressions - Add & Subtract 



Objective: Add and subtract rational expressions with and without 
common denominators. 

Adding and subtracting rational expressions is identical to adding and subtracting 
with integers. Recall that when adding with a common denominator we add the 
numerators and keep the denominator. This is the same process used with 
rational expressions. Remember to reduce, if possible, your final answer. 



Example 346. 



Same denominator, add numerators, combine like terms 





x-A 


x + 8 


X 2 


-2x- 


-8 x 2 -2x-8 

2x + 4 




x 2 -2x-% 






2(rc + 2) 




(x + 2)(x-4) 






2 



x-A 



Factor numerator and denominator 



Divide out (x + 2) 



Our Solution 



257 



Subtraction with common denominator follows the same pattern, though the sub- 
traction can cause problems if we are not careful with it. To avoid sign errors we 
will first distribute the subtraction through the numerator. Then we can treat it 
like an addition problem. This process is the same as "add the opposite" we saw 
when subtracting with negatives. 



Example 347. 

6x-12 15x-6 



3x — 6 32 — 6 



Add the opposite of the second fraction (distribute negative) 



f) r p 1 2 1 ^S t -\- o 

1 Add numerators, combine like terms 

32 — 6 32 — 6 

— Q<£ — Q 

Factor numerator and denominator 



32 — 6 

-3(32 + 2) 
3(2-2) 



Divide out common factor of 3 



— Our Solution 

2-2 



World View Note: The Rhind papyrus of Egypt from 1650 BC gives some of 
the earliest known symbols for addition and subtraction, a pair of legs walking in 
the direction one reads for addition, and a pair of legs walking in the opposite 
direction for subtraction.. 

When we don't have a common denominator we will have to find the least 
common denominator (LCD) and build up each fraction so the denominators 
match. The following example shows this process with integers. 



Example 348. 



% + \ The LCD is 12. Build up, multiply 6 by 2 and 4 by 3 
6 4 



|)| + I(| I Multip.y 



10 3 

— + — Add numerators 



258 



13 

— Our Solution 



The same process is used with variables. 
Example 349. 



1-7 A 7 

+ — — j The LCD is 6a 2 6 4 . We will then buildup each fraction 



3a 2 b 6a 6 4 



I - r? I — ^r + — rrl — I Multiply first fraction by 2b 3 and second by 
\2b 6 J 6a 2 b 6oo 4 Va/ 

14a6 3 4ofe ... ... . . 

+ Add numerators, no like terms to combine 



a 



Qa 2 b A 6a 2 b 4 

Uab 3 + 4ab 
6aW 

2a6(7fe 3 + 2) 
6aW 

7b 3 + 2 
3a b 3 



Factor numerator 



Reduce, dividing out factors 2, a, and b 



Our Solution 



The same process can be used for subtraction, we will simply add the first step of 
adding the opposite. 

Example 350. 

— -^ Add the opposite 

5a 4a 2 



A 7k 

- — h - LCD is 20a 2 . Build up denominators 
5a 4a 2 

Multiply first fraction by 4a , second by 5 
Our Solution 



4a \ 4 -76/5 

4a / 5a 4a 2 \5 

16a -356 
20a 2 



If our denominators have more than one term in them we will need to factor first 
to find the LCD. Then we build up each denominator using the factors that are 

259 



missing on each fraction. 
Example 351. 



+ — — Factor denominators to find LCD 



8a + 4 8 

4(2a+l) 8 LCD is 8 (2a + 1), build up each fraction 

77 )-T777 tt + vt-I 77 t I Multiply first fraction by 2, second by 2a + 1 

2/4(2a + l) 8V2a + ly iJ J ' J 



12 6a 2 + 3a 

f- 77777 7v Add numerators 



8(2o+l) 8(2a + l 
6a 2 + 3a + 12 



Our Solution 



8(2o+l) 

With subtraction remember to add the opposite. 



Example 352. 

x+1 x + 1 



x — 4 x 2 — 7x + 12 



Add the opposite (distribute negative) 



-H — = — Factor denominators to find LCD 

x — 4 x 2 — 7x + 12 

x — 4 (x — 4)(x — 3) LCD is (x — 4) (x — 3), build up each fraction 

x — 3 \ x ~\~ 1 — x — 1 

I — ^ — = 777- Only first fraction needs to be multiplied by x — 3 



x — 3/x — 4 x 2 — 7x + 12 
x 2 — 2x — 3 — x — 1 



t~ 7 ttw jt Add numerators, combine like terms 



(x-3)(x-4) (x-3)(x-4) 

x 2 - 3x - 4 
(x-3)(x-4) 

(x-4)(x + l) 

(x-3)(x-4) 

x + 1 
x — 3 



Factor numerator 



Divide out x — 4 factor 



Our Solution 



260 



7.4 Practice - Add and Subtract 



Add or subtract the rational expressions. Simplify your answers 
whenever possible. 



± J 


a + 3 a + 3 


3) 


t 2 + At 2t-7 


t-1 t-l 


5) 


2x 2 + 3 x 2 - 5x + 9 
x 2 — 6x + 5 x 2 — 6a; + 5 


7) 


5 5 
6r 8r 


9) 


8 5 

9*3 + ~^fi 


11) 


a+2 a-A 


2 A 


13) 


x-1 2x + 3 
Ax x 


15) 


5x + 3y 3x + 4j/ 
2x 2 y xy 2 


17) 


2z 3z 


Z-\ 2 + 1 


19) 


8 3 
x 2 -A x + 2 


21) 


t 5 


t - 3 At - 12 


23) 


2 4 


5x 2 + 5x 3x + 3 


25) 


t y 
y-t y+t 


27) 


x 2 


x 2 + 5x + 6 x 2 + 3x + 2 


29) 


x 7 


x 2 + 15x + 56 x 2 + 13a: + 42 


31) 


5x 18 
x 2 — x — 6 x 2 — 9 


33) 


2x 4 


x 2 -l x 2 + 2x-3 


35) 


x + 1 , x +6 


x 2 - 2x - 35 x 2 + 7x + 10 


37) 


4 - a 2 a - 2 


a 2 - 9 3 - a 


39) 


2z , 3z 3 


1-22 ' 2z + 1 Az 2 - 1 


41) 


2x - 3 3x - 1 


x 2 + 3x + 2 x 2 + 5x + 6 


43) 


2x + 7 3x - 2 





6x-8 



x-2 x-2 



4) 
6) 
8) 
10 
12 
14 
16 
18 
20 
22 
24 
26 
28 
30 
32 
34 
36 
38 
40 
42 
44 



a 2 + 3a 4 
a 2 + 5a — 6 a 2 + 5a - 

x x z 
7 3 



x + 5 x — 3 

~~ 8 12~~ 

2a -1 5a + 1 

3a 2 "" 9a~~ 

2c -d c + d 



c 2 d 
2 



zd 2 



X — 1 X + 1 

2 , 3 



x — 5 4x 

4x x 

x 2 - 25 x + 5 

-?- + 4 

x + 3 ' (x + 3) 2 

3a , 9a 



4a - 20 6a - 30 
x x — 5 



x — 5 

2x 



x 2 — 1 x 2 + 5x + 4 



2./: 



+ 



x 2 — 9 x 2 + x — 6 
4x 3 



x 2 — 2x — 3 x 2 — 5x + 6 
x — 1 , x + 5 



x 2 + 3x + 2 x 2 + 4x + 3 

3x + 2 x 
3x + 6 4-x 2 

4y 2 2 



y 2 - 1 1/ J/ + 1 
2r 1 



x + 2 4x + 5 



x 2 - 4x + 3 x 2 + 4x - 5 
3x - 8 , 2x - 3 



x 2 + 6x + 8 x 2 + 3x + 2 



261 



7.5 

Rational Expressions - Complex Fractions 

Objective: Simplify complex fractions by multiplying each term by the 
least common denominator. 

Complex fractions have fractions in either the numerator, or denominator, or usu- 
ally both. These fractions can be simplified in one of two ways. This will be illus- 
trated first with integers, then we will consider how the process can be expanded 
to include expressions with variables. 

The first method uses order of operations to simplify the numerator and denomi- 
nator first, then divide the two resulting fractions by multiplying by the recip- 
rocal. 

Example 353. 



2 1 

3 ~~ 4 



5 

6 "*" 2 

8 __ 3 
12 _ 12 

5 3 

6 "*" 6 



Y Get common denominator in top and bottom fractions 



Add and subtract fractions, reducing solutions 



5 

X- To divide fractions we multiply by the reciprocal 



12 JU 

m 



_5_ 

16 



Reduce 



Multiply 



Our Solution 



The process above works just fine to simplify, but between getting common 
denominators, taking reciprocals, and reducing, it can be a very involved process. 
Generally we prefer a different method, to multiply the numerator and denomi- 
nator of the large fraction (in effect each term in the complex fraction) by the 
least common denominator (LCD). This will allow us to reduce and clear the 
small fractions. We will simplify the same problem using this second method. 



Example 354. 



LCD is 12, multiply each term 



262 



2(12) 1(12) 

Reduce each fraction 



5(12) 1(12) 
6 ' 2 

2(4) -1(3) 
5(2) + 1(6) 

8-3 
10 + 6 

5 



Multiply 



Add and subtract 



Our Solution 
16 

Clearly the second method is a much cleaner and faster method to arrive at our 
solution. It is the method we will use when simplifying with variables as well. We 
will first find the LCD of the small fractions, and multiply each term by this LCD 
so we can clear the small fractions and simplify. 

Example 355. 

y- Identify LCD (use highest exponent) 

X 

LCD = a; 2 Multiply each term by LCD 

Reduce fractions (subtract exponents) 

Multiply 

Factor 



l(x 2 )- 


l(x 2 ) 

X 2 


l(x 2 )- 


1(* 2 ) 

X 


l(x 2 )-l 


l(a 


• 2 ) — X 




x 2 -l 




x 2 -x 


(x+l)( 


x-l) 


x(x - 


-1) 




x+1 



Divide out (x — 1) factor 



Our Solution 



x 



The process is the same if the LCD is a binomial, we will need to distribute 

-? 2 

' + ! Multiply each term by LCD , (x + 4) 



5' 2 



x + 4 



263 



3(x + 4) 

x + 4 



2(x + 4) 



A) | 2(x + 4) 


l > ' x + 4 


3-2(x + 4) 


5(x + 4)+2 


3-2rc-8 


5^ + 20 + 2 


-2a; -5 



Reduce fractions 



Distribute 



Combine like terms 



Our Solution 



5x + 22 
The more fractions we have in our problem, the more we repeat the same process. 



Example 356. 



2 _ 3 1 

ab 2 ab a ab 



-^r + ab 

a z b ab 



Y~ Idenfity LCD (highest exponents) 



LCD = a 2 b i Multiply each term by LCD 



2(q 2 b 3 ) 
ab 2 



3(a 2 6 3 ) l(a 2 6 3 ) 



ab 3 



ab 



^ + ab(aW) 



l(a 2 b 3 ) 
ab 



Reduce each fraction (subtract exponents) 



2a b — 3a + a b 2 
4b 2 + a 3 b 4 - ab 2 



Our Solution 



World View Note: Sophie Germain is one of the most famous women in mathe- 
matics, many primes, which are important to finding an LCD, carry her name. 
Germain primes are prime numbers where one more than double the prime 
number is also prime, for example 3 is prime and so is 2 • 3 + 1 = 7 prime. The 
largest known Germain prime (at the time of printing) is 183027- 2 265440 — 1 which 
has 79911 digits! 

Some problems may require us to FOIL as we simplify. To avoid sign errors, if 
there is a binomial in the numerator, we will first distribute the negative through 
the numerator. 

Example 357. 



Distribute the subtraction to numerator 



Identify LCD 



264 



LCD = (x + 3) (x - 3) Multiply each term by LCD 



(x-3)(x + 3)(x-3) (-x-3)(x + 3)(x-3) 

x + 3 x — 3 

(a; -3) (a: + 3) (a; -3) (x + 3)(x + 3)(x - 3) 

x + 3 x — 3 

(x - 3)(x - 3) + ( - x - 3)(x + 3) 
(x -3)(x-3) + (x + 3)(x + 3) 

x 2 — 6x + 9 — x 2 — 6x — 9 
x 2 — Qx + 9 + x 2 + 6x — 9 

-12a; 



2x 2 +18 


-12a; 


2(x 2 + 9) 


— 6x 



x 2 -9 



Reduce fractions 



FOIL 



Combine like terms 



Factor out 2 in denominator 



Divide out common factor 2 



Our Solution 



If there are negative exponents in an expression we will have to first convert these 
negative exponents into fractions. Remember, the exponent is only on the factor 
it is attached to, not the whole term. 



Example 358. 

m~ 2 + 2m _1 
m + 4m~ 2 

m z m 



. 4 



l(m 2 ) 2(m 2 ) 

m z m 

4(m 2 ) 



m(m 2 ) + 



l + 2m 
m 3 + 4 



Make each negative exponent into a fraction 
Multiply each term by LCD, m 2 

Reduce the fractions 
Our Solution 



Once we convert each negative exponent into a fraction, the problem solves 
exactly like the other complex fraction problems. 



265 



7.5 Practice - Complex Fractions 



Solve. 



1) 


1 + - 

X 


1-^2 


3) 


a-2 


4 

— a 

a 




1 1 


5) 


a 2 a 


1 , 1 



2- 4 



1 ) r 10 



x + 2 
3 



9)^ 

2a -3 



f2 



11) ^- 



x + i 



fi 



3 

13) -f 



., - \ 4a 2 b 

15 - ; z 



16ab 2 

-, _ 3 _ 10 

17)— 1^4 



x 



2) 



y 



4)i 



1 + i 



25 

a 



5 + a 



6) 


- + - 

6^2 
4 

6 2 -l 




8) 


4 +2 12 
2x - 


3 


5+ 2 15 

2x - 


3 


10) 


-5 
6-5 


•3 


10 + 

6-5 ' 


6 




2a 


3 


12) 


a-1 


a 


-6 
a-1 


-4 




X 




14) 


3x-2 





9x 2 -4 



16) 


1_I_A 

x X 2 


x X 2 


18) 


15 2 .. 
a; 2 x 


4 5 , . 



1 a; ' ™2 x 2 a; 



2x -, 12 



19) ^ 20) ^ 



3x - 4 3x + 10 



1 i 2 _ 18 

X — 1 + t X — 5 



21) *~e* 22) I 2 

x + 3H X + 7 + — — 

x — 4 x + 2 



266 





x 1 1 9 


23) 


X L ' 2x + 3 


r -1-3 5 




X ' d 2x + 3 


25) 


2 5 

6 6 + 3 

3 3 

6 + 6 + 3 




2 5 3 




j>2 ab a 2 



6 2 ^ ab ^ a 2 



29) y j/ 



24) 


a a — 2 
2 + 5 
a a — 2 




112 


26) 


J/ 2 ij/ x 2 


1 3 2 




j/ 2 zj/ X 2 




X — 1 £ + 1 


28) 


x + 1 a; — 1 


x — 1 x + 1 




x + 1 x — 1 




X + 1 1 — X 


30) 


X — 1 1 + X 


1 , 1 



y + 2 y-2 (x + 1)^ (x - 1)^ 



Simplify each of the following fractional expressions. 



o 9 O 9 

x — y x ii + xii 
31) _!_/_! 32) -^ V 



x 3 t/ — xy 3 4 — Ax x + x 2 

33) 1 V- 34) 5 

x~ z —y~ z A — x~ z 



x 



2 -6a;- 1 + 9 „„, £- 3 +y 



3 i .,-3 



35) -r-s 36 ) 



x 2 -9 y x 2 -x ^ ^y 



2 /v. — 1^,-1 I /). — 2 



267 



7.6 

Rational Expressions - Proportions 



Objective: Solve proportions using the cross product and use propor- 
tions to solve application problems 

When two fractions are equal, they are called a proportion. This definition can be 
generalized to two equal rational expressions. Proportions have an important 
property called the cross-product. 

a c 
Cross Product: If — = — then ad = be 
b a 

The cross product tells us we can multiply diagonally to get an equation with no 
fractions that we can solve. 



Example 359. 



— r = — Calculate cross product 
6 9 



268 



(20) (9) = 63 Multiply 

180 = 62 Divide both sides by 6 
"6" ~6~ 

30 = x Our Solution 



World View Note: The first clear definition of a proportion and the notation 
for a proportion came from the German Leibniz who wrote, "I write dy: x = dt: a; 
for dy is to x as dt is to a, is indeed the same as, dy divided by x is equal to dt 
divided by a. From this equation follow then all the rules of proportion." 

If the proportion has more than one term in either numerator or denominator, we 
will have to distribute while calculating the cross product. 



Example 360. 



x + 3 2 



Calculate cross product 



4 5 
5(x + 3) = (4) (2) Multiply and distribute 

5:c + 15 = 8 Solve 

— 15 — 15 Subtract 15 from both sides 

5x = — 7 Divide both sides by 5 

~5~ T~ 

7 

x = Our Solution 

5 



This same idea can be seen when the variable appears in several parts of the pro- 
portion. 



Example 361. 








4 _ 6 


Calculate cross product 




x 3 x + 2 




4(3x + 2) =6x 


Distribute 




122 + 8 = 62 


Move variables to one side 




-122 -122 


Subtract 122 from both sides 




8 = -62 


Divide both sides by — 6 




^6 ^6 






4 
~3 =X 


Our Solution 



269 



Example 362. 

2x -3 



5(2x- 


7x + 4 5 
3) = 2(7x + 4) 


10a; 


-15 = 14a; + 8 


-10a; 


-10a; 




-15 = 4a; + 8 




-8 -8 



Calculate cross product 

Distribute 

Move variables to one side 
Subtract 10a; from both sides 
Subtract 8 from both sides 



— 23 = 4x Divide both sides by 4 

~4~ ~4~ 

23 
— — = x Our Solution 
4 

As we solve proportions we may end up with a quadratic that we will have to 
solve. We can solve this quadratic in the same way we solved quadratics in the 
past, either factoring, completing the square or the quadratic formula. As with 
solving quadratics before, we will generally end up with two solutions. 



Example 363. 




k + 3 8 


Calculate cross product 


3 fc-2 


(fc + 3)(fc-2) = (8)(3) 


FOIL and multiply 


k 2 + k -6 = 24 


Make equation equal zero 


- 24 - 24 


Subtract 24 from both sides 


k 2 + k -30 = 


Factor 


(fc + 6)(fc-5) = 


Set each factor equal to zero 


A; + 6 = or fc-5 = 


Solve each equation 


-6-6 +5=5 


Add or subtract 


k = — 6 or k = 5 


Our Solutions 



Proportions are very useful in how they can be used in many different types of 
applications. We can use them to compare different quantities and make conclu- 
sions about how quantities are related. As we set up these problems it is impor- 
tant to remember to stay organized, if we are comparing dogs and cats, and the 
number of dogs is in the numerator of the first fraction, then the numerator of the 
second fraction should also refer to the dogs. This consistency of the numerator 
and denominator is essential in setting up our proportions. 

Example 364. 

A six foot tall man casts a shadow that is 3.5 feet long. If the shadow of a flag 
pole is 8 feet long, how tall is the flag pole? 

— We will put shadows in numerator, heights in denomintor 

height 



270 



3.5 

— — The man has a shadow of 3.5 feet and a height of 6 feet 
6 



— The flagpole has a shadow of 8 feet, but we don't know the height 
x 

—— = — This gives us our proportion, calculate cross product 
6 x 

3.5x= (8) (6) Multiply 

3.5a; = 48 Divide both sides by 3.5 
33 33 

x = 13.7ft Our Solution 



Example 365. 

In a basketball game, the home team was down by 9 points at the end of the 
game. They only scored 6 points for every 7 points the visiting team scored. 
What was the final score of the game? 

We will put home in numerator, visitor in denominator 



visiter 
x-9 



x 

6 
7 

x — 9 6 



Don't know visitor score, but home is 9 points less 
Home team scored 6 for every 7 the visitor scored 
This gives our proportion, calculate the cross product 



x 7 

7(x — 9) = 6x Distribute 

7x — 63 = 6x Move variables to one side 

-7x — 7x Subtract 7x from both sides 



— 63 = — x Divide both sides by — 1 

63 = x We used a: for the visitor score. 

63 — 9 = 54 Subtract 9 to get the home score 

63 to 54 Our Solution 



271 



7.6 Practice - Proportions 



Solve each proportion 


«T=i 




3)1 = 1 




5)!=! 




7)=fl= 


_ 8 

2 


01 2 _ 10 

a ' 9 p-4 


11) 6 " 7 10 


_ 6 
4 


13) l = x 


+ 2 
9 


15) 3 -- 


a 


ld J 10 , 


2+2 


7 t> + 6 


4 
~ 9 


19) 7 = 


4 


1J J x-l 


x — 6 


21) * + 5 = 


6 
x-2 


23) m + 3 


11 


"°l 4 


m — 4 


25) * = 

7 p + 4 


_ p + 5 
3 


2?) « + 4 = 


-3 
n-2 


29) -^r = 


x + 2 



2) - = - 

'9 6 



4)i = l 

7 x 8 



^ n-10 9 
6 )^=3 

o\ 8 3 



10) 
12) 
14) 
16) 
18) 
20) 
22) 
24) 
26) 
28) 
30) 



5 x-8 
9 3 



n + 2 9 
9 r 



4 r — 4 

n n — 4 



8 3 

x + 1 x + 2 



9 2 

n + 8 n-9 



10 4 

fc + 5 _ 8 
fc-6 ~~ 5 

4 _ x + 5 
x — 3 5 

x — 5 4 



8 x-l 

5 n — 4 



71 + 1 10 

1 n + 2 



n + 3 2 

x — 5 — 3 



4 x + 3 



Answer each question. Round your answer to the nearest tenth. Round 
dollar amounts to the nearest cent. 

31) The currency in Western Samoa is the Tala. The exchange rate is 

approximately $0.70 to 1 Tala. At this rate, how many dollars would you 
get if you exchanged 13.3 Tala? 

32) If you can buy one plantain for $0.49 then how many can you buy with 

$7.84? 



272 



33) Kali reduced the size of a painting to a height of 1.3 in. What is the new 

width if it was originally 5.2 in. tall and 10 in. wide? 

34) A model train has a scale of 1.2 in : 2.9 ft. If the model train is 5 in tall then 

how tall is the real train? 

35) A bird bath that is 5.3 ft tall casts a shadow that is 25.4 ft long. Find the 

length of the shadow that a 8.2 ft adult elephant casts. 

36) Victoria and Georgetown are 36.2 mi from each other. How far apart would 

the cities be on a map that has a scale of 0.9 in : 10.5 mi? 

37) The Vikings led the Timberwolves by 19 points at the half. If the Vikings 

scored 3 points for every 2 points the Timberwolves scored, what was the 
half time score? 

38) Sarah worked 10 more hours than Josh. If Sarah worked 7 hr for every 2 hr 
Josh worked, how long did they each work? 

39) Computer Services Inc. charges $8 more for a repair than Low Cost 

Computer Repair. If the ratio of the costs is 3 : 6, what will it cost for the 
repair at Low Cost Computer Repair? 

40) Kelsey's commute is 15 minutes longer than Christina's. If Christina drives 12 

minutes for every 17 minutes Kelsey drives, how long is each commute? 



273 



7.7 

Rational Expressions - Solving Rational Equations 



Objective: Solve rational equations by identifying and multiplying by 
the least common denominator. 

When solving equations that are made up of rational expressions we will solve 
them using the same strategy we used to solve linear equations with fractions. 
When we solved problems like the next example, we cleared the fraction by multi- 
plying by the least common denominator (LCD) 



Example 366. 



2 5 3 
-x — — = — Multiply each term by LCD, 12 



3 6 4 



2(12) 5(12) 3(12) 

— — -x — - = — — - Reduce tractions 

3 6 4 



2(4)x-5(2) = 3(3) Multiply 

8x - 10 = 9 Solve 

+ 10 + 10 AddlOtoboth sides 

8a: = 19 Divide both sides by! 



19 

x = — Our Solution 

8 



We will use the same process to solve rational equations, the only difference is our 



274 



LCD will be more involved. We will also have to be aware of domain issues. If our 
LCD equals zero, the solution is undefined. We will always check our solutions in 
the LCD as we may have to remove a solution from our solution set. 



Example 367. 






5x + 5 _ x 

— + 3a; = - 

x+2 x+2 


(5x + 5)(x + 2) 
x + 2 


+ 3x(x + 2) = x2( - X \ 2) 
x + 2 




5x + 5 + 3a;(a; + 2) = a; 2 




5x + 5 + 3a: 2 + 6x = x 2 




3a: 2 + llx + 5 = a: 2 




rp^ rp^ 



Multiply each term by LCD, (x + 2) 

Reduce fractions 

Distribute 
Combine like terms 
Make equation equal zero 
Subtract x 2 from both sides 
2a: 2 + lla: + 5 = Factor 
(2x + 1) (x + 5) = Set each factor equal to zero 
2a; +1 = or a: + 5 = Solve each equation 
-1-1 -5-5 



2x = — 1 or x = — 5 
~2 ~2 

x = or — 5 Check solutions, LCD can't be zero 

2 

1 3 

— + 2 = — — 5 + 2 = — 3 Neither make LCD zero, both are solutions 

x = — — or — 5 Our Solution 



The LCD can be several factors in these problems. As the LCD gets more com- 
plex, it is important to remember the process we are using to solve is still the 
same. 



Example 


368. 










X 


1 

1 x + l 


5 






x + 2 


(x+ l)(x 


+ 2) 


x(x + l)(x 
x + 2 


+ 2) ! l(a 


: + l)(ar + 2) 
x + l 


_5(x + l)(: 

(x + l) (a: 


c + 2) 
^ + 2) 



Multiply terms by LCD, (x + l)(x + 2) 



Reduce fractions 



275 



(-3 + 





x(x + l) + l(x + 2) = 


= 5 




x 2 + x + x + 2 = 


= 5 




x 2 + 2x + 2 = 


= 5 




-5- 


-5 




x 2 + 2x - 3 = 


= 




(x + 3)(x-l) = 


= 




x + 3 = or x — 1 = 


= 




-3-3 +1+1 




£ = — 3 or x = 


= 1 


)( 


-3 + 2) = (-2)(-l) = 


--2 




(l + l)(l + 2) = (2)(3) = 


= 6 




x = — 3 or 


1 



Distribute 

Combine like terms 

Make equatino equal zero 

Subtract 6 from both sides 

Factor 

Set each factor equal to zero 

Solve each equation 

Check solutions, LCD can't be zero 
Check — 3 in (x + 1) (x + 2) , it works 
Check 1 in (x + 1) (x + 2) , it works 
Our Solution 



In the previous example the denominators were factored for us. More often we 
will need to factor before finding the LCD 



Example 369. 



x 



x — 1 



11 



Factor denominator 



x-2 x 2 -3x + 2 
(x-l)(x-2) 

LCD = (x-l)(x-2) Identify LCD 



x{x - l)(x - 2) _ \(x - l)(x - 2) _ ll(a; - l)(x - 2) 
x-l x-2 (x- l)(x-2) 



Multiply each term by LCD, reduce 



x(x-2)-l(x-l) = ll Distribute 

x 2 — 2x — x + 1 = 1 1 Combine like terms 

x 2 — 3x + 1 = 1 1 Make equation equal zero 



11 — 11 Subtract 1 1 from both sides 



x 



3x — 10 = Factor 



(x — 5) (x + 2) = Set each factor equal to zero 

x — 5 = or x + 2 = Solve each equation 
+5+5 -2-2 

x = 5 or x = — 2 Check answers, LCD can't be 

(5 - 1)(5 - 2) = (4) (3) = 12 Check 5 in (x - l)(x - 2), it works 

( - 2 - 1)( - 2 - 2) = ( - 3)( - 4) = 12 Check - 2 in (x - l)(x - 2), it works 



276 



1 = 5 or — 2 Our Solution 

World View Note: Maria Agnesi was the first women to publish a math text- 
book in 1748, it took her over 10 years to write! This textbook covered everything 
from arithmetic thorugh differential equations and was over 1,000 pages! 

If we are subtracting a fraction in the problem, it may be easier to avoid a future 
sign error by first distributing the negative through the numerator. 

Example 370. 

x-2 x + 2 5 



x-3 x+2 8 
x-2 -x-2 5 



x — 3 x + 2 



Distribute negative through numerator 

Identify LCD, 8(x — 3) (x + 2), multiply each term 



(x-2)8(x-3)(x + 2) (- x -2)8(x-3)(x + 2) _ 5 -8(x -3)(x + 2) ^^ 
x-3 ' x+2 8 



8{x - 2)(x + 2) +8( - x - 2)(x - 3) = 5(x - 3){x + 2) FOIL 

3(a; 2 - 4) + 8( - x 2 + x + 6) = 5(x 2 - x - 6) Distribute 

8x 2 - 32 - 8x 2 + 8x + 48 = 5x 2 - 5x - 30 Combine like terms 

8a; + 16 = 5x 2 — 5x — 30 Make equation equal zero 

— 8x — 16 — 8a; — 16 Subtract 8a; and 16 

= 5x 2 - 13a; - 46 Factor 

= (5a; — 23) (x + 2) Set each factor equal to zero 

5x — 23 = or x + 2 = Solve each equation 
+ 23 + 23 -2-2 



5a; = 23 or x = — 2 

23 
x= — or —2 Check solutions, LCD can't be 

5 



(f-3)(f + 2)=8(|)(f)=^ Check f in 8(x - 3) (x + 2), it works 



\( - 2 - 3) ( - 2 + 2) = 8( - 5) (0) = Check - 2 in 8(x - 3) (x + 2) , can't be 0! 

23 

5 



23 

x = —— Our Solution 



In the previous example, one of the solutions we found made the LCD zero. 
When this happens we ignore this result and only use the results that make the 
rational expressions defined. 



277 



7.7 Practice - Solving Rational Equations 



Solve the following equations for the given variable: 



1) 3x-^-- = 

' 2 x 



3) a: + _?° T = _*!L_2 



5) x 

7) 

9) 

11) 
13) 

15) 



x -4 x — 4 
6 2x 



x — 3 x — 3 

2x 4x + 5 3 



3x - 4 6x - 1 3x - 4 
3m 7 3 



2m - 5 3m + 1 2 
4-x 12 



1 — x 3 — x 

_J_ 1 = V-2 

y-3 2 y-4 

1 1 3x + 8 



x + 2 2-x x 2 - 4 



1_x x + 1 x — 1 5 

' x-1 ~~ x+1 ~~ 6^ 

ln x 3 , 2x + l -, 8x 2 

iy ) ^TT + T^^- 1 



4x 2 -l 



21: 



x-2 1 



x + 3 x — 2 x 2 + x — 6 



t-\n\ 3 j_ x ~ 1 5x + 20 

' x + 2 "*" x + 5 ~~ 6x + 24 

25) _JL 2 _ = 4x2 

' X — 1 X + 1 X 2 — 1 

2x 3 - 8x 2 



2 7)iTT 



x + 5 x 2 + 6x + 5 



-4^ 



OQ") £: 5 x + 3 

' x-9 x-3 x 2 -12x + 27 

/ x-e^x + S - x 2 -3x-18 



oo\ 4x + l 5x-3_ 8x 2 

' x + 3 x-1 ~~ x 2 + 2x - 3 



2) 


x + l = 




x + 1 


4) 


x 2 + 6 
x-1 ~ l ~ 


x ~ 2 -2x 

x — 1 


6) 


x- 4 


12 +1 


x- 1 S 


-x ' X 


8) 


6x + 5 


2 3x 


2x 2 -2x 


1 — X 2 X 2 — 1 


10 


. 4x 


4 1 


' 2x-6 


5x-15 2 


12 


)/ + 
3 — x 


1 _ 3 

2 4-x 


14 


2 
3 — x 


6 "I 

8-x 


16 


( x + 2 


1 3x-3 


' 3x-l 


x 3x 2 — x 


18 


) X_1 + 
x — 3 


x + 2 _ 3 
x + 3 4 


20 


( 3x — 5 

5x — 5 


5x — 1 x — 4 ^ 
7x - 7 1 - x 


22 


1 x-2 ' 


x + 4 1 


2x + 1 2x 2 - 3x - 2 


24 


\ x 
' x + 3 


4 -5x 2 


x — 2 x 2 + x — 6 


26 


2x 

' x + 2 ' 


2 3x 


x - 4 x 2 - 2x - 8 


28 


. X 

' x+1 


3 -2x 2 


x + 3 x 2 + 4x + 3 


30 


. x — 3 

' x + 6 ~ l ~ 


x-2 x 2 


x-3 x 2 + 3x-18 


32 


. x + 3 


x-2 9x 2 


x + 1 x 2 — x — 2 


34 


( 3x-l 


2x-3 -3x 2 


' x + 6 


x-3 x 2 + 3x - 18 



278 



7.8 

Rational Expressions - Dimensional Analysis 



Objective: Use dimensional analysis to preform single unit, dual unit, 
square unit, and cubed unit conversions. 

One application of rational expressions deals with converting units. When we con- 
vert units of measure we can do so by multiplying several fractions together in a 
process known as dimensional analysis. The trick will be to decide what fractions 
to multiply. When multiplying, if we multiply by 1, the value of the expression 
does not change. One written as a fraction can look like many different things as 
long as the numerator and denominator are identical in value. Notice the numer- 
ator and denominator are not identical in appearance, but rather identical in 
value. Below are several fractions, each equal to one where numerator and denom- 
inator are identical in value. 



1 _ 4 _ 2 _ 100cm _ lib lhr _60min 

14 1 lm 16oz 60min lhr 



The last few fractions that include units are called conversion factors. We can 
make a conversion factor out of any two measurements that represent the same 
distance. For example, 1 mile = 5280 feet. We could then make a conversion 
factor because both values are the same, the fraction is still equal to one. 

Similarly we could make a conversion factor — — — . The trick for conversions will 

17 lmi 



279 



be to use the correct fractions. 

The idea behind dimensional analysis is we will multiply by a fraction in such a 
way that the units we don't want will divide out of the problem. We found out 
when multiplying rational expressions that if a variable appears in the numerator 
and denominator we can divide it out of the expression. It is the same with units. 
Consider the following conversion. 



Example 371. 



17.37 miles to feet Write 17.37 miles as a fraction, put it over 1 

To divide out the miles we need miles in the denominator 



17.37mi V ??ft 



1 J\??mi 
17.37miV 5280ft \ 



1 J\ lmi J 

17.37 Y 5280ft \ 



1 J\ 1 J 

91,713.6ft Our Solution 



We are converting to feet , so this will go in the numerator 
Fill in the relationship described above, 1 mile = 5280 feet 
Divide out the miles and multiply across 



In the previous example, we had to use the conversion factor 



would divide out. If we had used 



lmi 
5280 ft 



5280 ft 
lmi 



so the miles 



we would not have been able to divide out 



the miles. This is why when doing dimensional analysis it is very important to 
use units in the set-up of the problem, so we know how to correctly set up the 
conversion factor. 



Example 372. 

If 1 pound = 16 ounces, how many pounds 435 ounces? 



1 



435oz 



435oz 



435 oz V??lbs 



??oz 



libs 
16oz 



Write 435 as a fraction, put it over 1 

To divide out oz, 

put it in the denominator and lbs in numerator 

Fill in the given relationship, 1 pound = 16 ounces 



280 



435Vllbs\ 435 lbs 

Divide out oz, multiply across. Divide result 



i A i6 y i6 

27.1875 lbs Our Solution 



The same process can be used to convert problems with several units in them. 
Consider the following example. 



Example 373. 

A student averaged 45 miles per hour on a trip. What was the student's speed in 
feet per second? 



45 mi 
hr 



45miV 5280 ft \ 
hr J\ lmi / 



"per" is the fraction bar, put hr in denominator 

To clear mi they must go in denominator and become ft 



45miV5280fty^_\ Todearhrtheymustgoinnlimeratorandbecomesec 
hr J\ lmi J\ 3600 secy 



45 V 5280ft V 1 

1 A 1 A 3600 sec 



Divide out mi and hr. Multiply across 



237600ft _. . . 

— — — Divide numbers 

3600 sec 

66 ft per sec Our Solution 

If the units are two-dimensional (such as square inches - in 2 ) or three-dimensional 
(such as cubic feet - ft 3 ) we will need to put the same exponent on the conversion 
factor. So if we are converting square inches (in 2 ) to square ft (ft 2 ), the conversion 

factor would be squared, ( —rr- ) . Similarly if the units are cubed, we will cube 
the convesion factor. 

Example 374. 

Convert 8 cubic feet to yd 3 Write 8ft 3 as fraction, put it over 1 

5 ft 3 ", 

To clear ft, put them in denominator, yard in numerator 



281 



!ft 3 \/ ??vd \ 3 

-, — II 77W7 I Because the units are cubed, 
1 )\ ??ft J 



we cube the conversion factor 

3 



, r, 3 \ / i jX'J 

— — )l — — ) Evaluate exponent, cubing all numbers and units 



^) — 3 



lyd 3 \ 8yd 3 ..... 

1 - Multiply across and divide 



i yv 27 y 27 

0.296296yd 3 Our Solution 



When calculating area or volume, be sure to use the units and multiply them as 
well. 



Example 375. 

A room is 10 ft by 12 ft. How many square yards are in the room? 

A = lw = (10ft) (12 ft) = 120ft 2 Multiply length by width, also multiply units 

120ft 2 ', w ., , . 

Write area as o traction, put it over 1 



'' ''' ' Put ft in denominator to clear, 



! A ??ft , 

square conversion factor 

120ft 2 \( lyd * " 



1 \ 3ft 



Evaluate exponent, squaring all numbers and units 



/ 120ft 2 \/lyd 2 \ _.. . . _, 



120V!yd 2 ^ 120yd 2 .. ... . ..... 

" ' — Multiply across and divide 



1 )\ 9 J 9 

13.33yd 2 Our solution 

282 



To focus on the process of conversions, a conversion sheet has been included at 
the end of this lesson which includes several conversion factors for length, volume, 
mass and time in both English and Metric units. 

The process of dimensional analysis can be used to convert other types of units as 
well. If we can identify relationships that represent the same value we can make 
them into a conversion factor. 



Example 376. 

A child is perscribed a dosage of 12 mg of a certain drug and is allowed to refill 
his prescription twice. If a there are 60 tablets in a prescription, and each tablet 
has 4 mg, how many doses are in the 3 prescriptions (original + 2 refills)? 

Convert 3 Rx to doses Identify what the problem is asking 
1 Rx = 60 tab, 1 tab = 4 mg, 1 dose = 12 mg Identify given conversion factors 



3Rx 



Write 3Rx as fraction, put over 1 



/ 3Rx\/ 60 tab 



V 1 A lRx 



Convert Rx to tab, put Rx in denominator 



3Rx 



60 tab V4mg 
lRx )\ ltab 



Convert tab to mg, put tab in denominator 



3 Rx V 60 tab V 4mg V 1 dose 
~T~ )\ lRx ;VltabyVl2mg 



Convert mg to dose, put mg in denominator 



— if — if — )( — t-z — ) Divide out Rx, tab, and mg, multiply across 



720 dose 
12 



Divide 



60 doses Our Solution 



World View Note: Only three countries in the world still use the English 
system commercially: Liberia (Western Africa), Myanmar (between India and 
Vietnam), and the USA. 



283 



Conversion Factors 



Leu 


Lgth 


English 


Metric 


12 in = 1 ft 


1000 mm = 1 m 


1 yd = 3 ft 


10 mm = 1 cm 


1 yd = 36 in 


100 cm = 1 m 


1 mi = 5280 ft 


10 dm = 1 m 




1 dam = 10 m 




1 hm = 100 m 




1 km = 1000 m 


English/Metric 


2.54 cm == 1 in 


1 m = 


3.28 ft 


1.61 km 


= 1 mi 



Volume 


English Metric 


1 c = 


8 oz 1 mL = 1 cc = 1 cm 3 


1 pt 


= 2 c 1 L = 1000 mL 


1 qt = 


= 2 pt 1 L = 100 cL 


1 gal = 


= 4 qt 1 L = 10 dL 




1000 L = 1 kL 




English/Metric 




16.39 mL = 1 in 3 




1.06 qt = 1 L 




3.79 L = lgal 



Area 



English 

1 ft 2 = 144 



m 



1 yd 2 = 9 ft 2 
1 acre = 43,560 ft 2 
640 acres = 1 mi 2 



Metric 

1 a = 100 m 2 
1 ha = 100 a 



English/Metric 

1 ha = 2.47 acres 



Weight (Mass) 



English 

1 lb = 16 oz 
1 T = 2000 lb 



Metric 

1 g = 1000 mg 

1 g = 100 eg 

1000 g = 1 kg 

1000 kg = 1 t 



English/Metric 

28.3 g = 1 oz 
2.2 lb = 1 kg 



Time 



60 sec = 1 min 

60 min = 1 hr 

3600 sec = 1 hr 

24 hr = 1 day 



284 



7.8 Practice - Dimensional Analysis 

Use dimensional analysis to convert the following: 

1) 7 mi. to yards 

2) 234 oz. to tons 

3) 11.2 mg to grams 

4) 1.35 km to centimeters 

5) 9,800,000 mm (milimeters) to miles 

6) 4.5 ft 2 to square yards 

7) 435,000 m 2 to sqaure kilometers 

8) 8 km 2 to square feet 

9) 0.0065 km 3 to cubic meters 

10) 14.62 in 3 to cubic centimeters 

11) 5,500 cm 3 to cubic yards 

12) 3.5 mph (miles per hour) to feet per second 

13) 185 yd. per min. to miles per hour 

14) 153 ft/s (feet per second) to miles per hour 

15) 248 mph to meters per second 

16) 186,000 mph to kilometers per year 

17) 7.50 T/yd 2 (tons per square yard) to pounds per square inch 

18) 16 ft/s 2 to kilometers per hour squared 



285 



Use dimensional analysis to solve the following: 

19) On a recent trip, Jan traveled 260 miles using 8 gallons of gas. How many 
miles per 1-gallon did she travel? How many yards per 1-ounce? 

20) A chair lift at the Divide ski resort in Cold Springs, WY is 4806 feet long and 
takes 9 minutes. What is the average speed in miles per hour? How many feet 
per second does the lift travel? 

21) A certain laser printer can print 12 pages per minute. Determine this printer's 
output in pages per day, and reams per month. (1 ream = 5000 pages) 

22) An average human heart beats 60 times per minute. If an average person lives 
to the age of 75, how many times does the average heart beat in a lifetime? 

23) Blood sugar levels are measured in miligrams of gluclose per deciliter of blood 
volume. If a person's blood sugar level measured 128 mg/dL, how much is 
this in grams per liter? 

24) You are buying carpet to cover a room that measures 38 ft by 40 ft. The 
carpet cost $18 per square yard. How much will the carpet cost? 

25) A car travels 14 miles in 15 minutes. How fast is it going in miles per hour? 
in meters per second? 

26) A cargo container is 50 ft long, 10 ft wide, and 8 ft tall. Find its volume in 
cubic yards and cubic meters. 

27) A local zoning ordinance says that a house's "footprint" (area of its ground 

floor) cannot occupy more than - of the lot it is built on. Suppose you own a 

l c 

- acre lot, what is the maximum allowed footprint for your house in square 

feet? in square inches? (1 acre = 43560 ft 2 ) 

28) Computer memory is measured in units of bytes, where one byte is enough 
memory to store one character (a letter in the alphabet or a number). How 
many typical pages of text can be stored on a 700-megabyte compact disc? 
Assume that one typical page of text contains 2000 characters. (1 megabyte = 
1,000,000 bytes) 

29) In April 1996, the Department of the Interior released a "spike flood" from the 
Glen Canyon Dam on the Colorado River. Its purpose was to restore the river 
and the habitants along its bank. The release from the dam lasted a week at a 
rate of 25,800 cubic feet of water per second. About how much water was 
released during the 1-week flood? 

30) The largest single rough diamond ever found, the Cullinan diamond, weighed 
3106 carats; how much does the diamond weigh in miligrams? in pounds? 

(1 carat - 0.2 grams) 



286 



Chapter 8 : Radicals 



8.1 Square Roots 288 

8.2 Higher Roots 292 

8.3 Adding Radicals 295 

8.4 Multiply and Divide Radicals 298 

8.5 Rationalize Denominators 303 

8.6 Rational Exponents 310 

8.7 Radicals of Mixed Index 314 

8 . 8 Complex Numbers 318 



287 



Radicals - Square Roots 



Objective: Simplify expressions with square roots. 

Square roots are the most common type of radical used. A square root "un- 
squares" a number. For example, because 5 2 = 25 we say the square root of 25 is 5. 
The square root of 25 is written as v25- 



World View Note: The radical sign, when first used was an R with a line 
through the tail, similar to our perscription symbol today. The R came from the 
latin, "radix", which can be translated as "source" or "foundation". It wasn't until 
the 1500s that our current symbol was first used in Germany (but even then it 
was just a check mark with no bar over the numbers! 

The following example gives several square roots: 



Example 377. 



VT = 1 


V121 = 11 


\/l = 2 


V625=25 


\/9 = 3 


V- 81= Undefined 



The final example, V '— 81 is currently undefined as negatives have no square root. 
This is because if we square a positive or a negative, the answer will be positive. 
Thus we can only take square roots of positive numbers. In another lesson we will 
define a method we can use to work with and evaluate negative square roots, but 
for now we will simply say they are undefined. 

Not all numbers have a nice even square root. For example, if we found v8 on 
our calculator, the answer would be 2.828427124746190097603377448419... and 
even this number is a rounded approximation of the square root. To be as accu- 
rate as possible, we will never use the calculator to find decimal approximations of 
square roots. Instead we will express roots in simplest radical form. We will do 
this using a property known as the product rule of radicals 

Product Rule of Square Roots: \J a • b = \fa • y/b 



We can use the product rule to simplify an expression such as \/36 • 5 by spliting 
it into two roots, v36 • Vo, and simplifying the first root, 6v5- The trick in this 



288 



process is being able to translate a problem like \/180 into \/36 • 5. There are sev- 
eral ways this can be done. The most common and, with a bit of practice, the 
fastest method, is to find perfect squares that divide evenly into the radicand, or 
number under the radical. This is shown in the next example. 



Example 378. 



75 75 is divisible by 25, a perfect square 

\/25 • 3 Split into factors 

25 • \/3 Product rule, take the square root of 25 

5v3 Our Solution 



If there is a coefficient in front of the radical to begin with, the problem merely 
becomes a big multiplication problem. 



Example 379. 



5v 63 63 is divisible by 9, a perfect square 

5\/9 • 7 Split into factors 

5 v9 • v7 Product rule, take the square root of 9 

5 • 3v7 Multiply coefficients 

15\/7 Our Solution 



As we simplify radicals using this method it is important to be sure our final 
answer can be simplified no more. 



Example 380. 



72 72 is divisible by 9, a perfect square 

v9 • 8 Split into factors 

\/9 • \/8 Product rule, take the square root of 9 

3\/8 But 8 is also divisible by a perfect square, 4 

3-\/4 • 2 Split into factors 

3v4 • v2 Product rule, take the square root of 4 

3 • 2\/2 Multiply 



289 



6\/2 Our Solution. 



The previous example could have been done in fewer steps if we had noticed that 
72 = 36 • 2, but often the time it takes to discover the larger perfect square is more 
than it would take to simplify in several steps. 

Variables often are part of the radicand as well. When taking the square roots of 
variables, we can divide the exponent by 2. For example, vx 8 = x 4 , because we 
divide the exponent of 8 by 2. This follows from the power of a power rule of 
expoennts, (a; 4 ) 2 = rr 8 . When squaring, we multiply the exponent by two, so when 
taking a square root we divide the exponent by 2. This is shown in the following 
example. 



Example 381. 

— 5 v 18x 4 y 6 z 10 18 is divisible by 9, o perfect square 
— 5 v 9 • 2x 4 y 6 z 10 Split into factors 

— 5v9 • v2 • v x 4 ■ yy & ■ vz 10 Product rule, simplify roots, divide exponents by 2 

— 5 • 3x 2 y 3 z 5 \/2 Multiply coefficients 
— 15x 2 y 3 z 5 \/2 Our Solution 



We can't always evenly divide the exponent on a variable by 2. Sometimes we 
have a remainder. If there is a remainder, this means the remainder is left inside 
the radical, and the whole number part is how many are outside the radical. This 
is shown in the following example. 



Example 382. 

V 20x 5 y 9 z 6 20 is divisible by 4, a perfect square 

\/4 • 5x 5 y 9 z 6 Split into factors 

vi • \/5 • v x 5 • a/ y 9 ■ vz 6 Simplify, divide exponents by 2, remainder is left inside 

2x 2 y 4 z 3 \/5xy Our Solution 



290 



8.1 Practice - Square Roots 



Simplify. 


1) \/245 


3) 


v^ 


5) 


\/T2 


7) 


3\/l2 


9) 6\/l28 


11 


) -8\/392 


13 


) Vl92n 


15 


) \/196d 2 


17 


) V252a; 2 


19 


) -VlOOk 4 


21 


) -7\/64x 4 


23 


) — 5\/36m 


25 


) i/45x 2 y 2 


27 


) -\/l6x 3 y 3 


29 


) y / 320a; 4 ?/ 4 


31 


) Gx/SOa;?/ 2 


33 


) 5v / 245xV 


35 


) -2Vl80« 3 t; 


37 


) -8^180x 4 y 2 z 4 


39 


) 2^80hj 4 k 


41 


) — 4y'54mnp 2 



2) 


V125 


4) Vl96 


6) 


\/72 


8) 


5\/32 


10 


) 7\/l28 


12 


) -7\/63 


14 


) V3436 


16 


) VlOOn 3 


18 


) \/200a 3 


20 


) -4^175p 4 


22 


) -2\/l28n 


24 


) 8\/ll2p 2 


26 


) \/72a 3 6 4 


28 


) V512a 4 6 2 


30 


) \/512m 4 n 3 


32 


) 8V98mn 


34 


) 2v / 72xV 


36 


) -by/72x 3 y 4 


38 


) 6V50a 4 6c 2 


40 


) -y/32xy 2 z 3 


42 


) -8v / 32m 2 p 4 g 



291 



8.2 



Radicals - Higher Roots 



Objective: Simplify radicals with an index greater than two. 

While square roots are the most common type of radical we work with, we can 
take higher roots of numbers as well: cube roots, fourth roots, fifth roots, etc. Fol- 
lowing is a definition of radicals. 

Va = b if b m = a 

The small letter m inside the radical is called the index. It tells us which root we 
are taking, or which power we are "un-doing". For square roots the index is 2. As 
this is the most common root, the two is not usually written. 

World View Note: The word for root comes from the French mathematician 
Franciscus Vieta in the late 16th century. 

The following example includes several higher roots. 



Example 383. 



\/125=5 


V-64 = -4 


t/8T = 3 


V- 128 = - 2 


v / 32 = 2 


V 7 — 16 = undefined 



We must be careful of a few things as we work with higher roots. First its impor- 
tant not to forget to check the index on the root. Vol = 9 but v81 = 3. This is 
because 9 2 = 81 and 3 4 = 81. Another thing to watch out for is negatives under 
roots. We can take an odd root of a negative number, because a negative number 
raised to an odd power is still negative. However, we cannot take an even root of 
a negative number, this we will say is undefined. In a later section we will discuss 
how to work with roots of negative, but for now we will simply say they are unde- 
fined. 

We can simplify higher roots in much the same way we simplified square roots, 
using the product property of radicals. 

Product Property of Radicals: r \/ab = r \fa • r \/b 

Often we are not as familiar with higher powers as we are with squares. It is 
important to remember what index we are working with as we try and work our 
way to the solution. 



292 



Example 384. 

o i 

V 54 We are working with a cubed root, want third powers 

2 3 = 8 Test 2, 2 3 = 8, 54 is not divisible by 8. 

3 3 = 27 Test 3, 3 3 = 27, 54 is divisible by 27! 

\/27- 2 Write as factors 

v27 • v2 Product rule, take cubed root of 27 

3 v2 Our Solution 

Just as with square roots, if we have a coefficient, we multiply the new coefficients 
together. 



We are working with a fourth root, want fourth powers 

Test 2, 2 4 = 16, 48 is divisible by 16! 

Write as factors 

Product rule, take fourth root of 16 

Multiply coefficients 

Our Solution 

We can also take higher roots of variables. As we do, we will divide the exponent 
on the variable by the index. Any whole answer is how many of that varible will 
come out of the square root. Any remainder is how many are left behind inside 
the square root. This is shown in the following examples. 

Example 386. 

^Jx 25 y 17 z 3 Divide each exponent by 5, whole number outside, remainder inside 
x 5 y 3 yy 2 z 3 Our Solution 



Example 


385. 




3a/48 




2 4 = 16 




3^16-3 


3 


VT6-V3 




3-2V3 




6V3 



25 _ C Z? fl O^ rJ0 



remain inside. For the y, we divided — = 3R2, so y 3 came out, and y 2 remains 
inside. For the z, when we divided T = 0f?3, all three or z 3 remained inside. The 

5 



In the previous example, for the x, we divided — = 5-RO, so x came out, no x's 

— = 3R2, so y 3 came r " if " wl " 2 

5 ' a 

| = 0R3, all th— — - 3 
following example includes integers in our problem. 

Example 387. 

2 v 40a 4 6 8 Looking for cubes that divide into 40. The number 8 works! 

2 v 8 • 5a 4 6 8 Take cube root of 8, dividing exponents on variables by 3 

2 • 2ab 2 V 5ab 2 Remainders are left in radical. Multiply coefficients 

4ab 2 V5ab 2 Our Solution 



293 



8.2 Practice - Higher Roots 



Simplify. 

I) V625 
3) V750 
5) V875 

7) -4V96 
9) 6\/TT2 

II) -Vll2 
13) " ' 



17; 
19; 
21; 

23; 

25; 
27; 
29; 

31' 
33; 

35; 
37; 
39; 
41 



V648a 2 



V224n 3 



V224p 5 

- 3\/896r 

- 2\/- 48w 7 



7\/320n 6 



V- 135xV 



V 3 ^ 



4 4 

If 



3 a/256xV 



7a/- 81zV 



2V375^V 



-3\/l92a6 2 
6%/- 54m 8 n 3 p 7 
6i/ 6A8x b y 7 z 2 
7\/l28h 6 j 8 k 8 



2) V375 


4) a/250 


6) 


V24 


8) 


-8a/48 


10 


) 3V48 


12 


) 5V243 


14 


) \/64n 3 


16 


) V- 96a; 3 


18 


) V256x 6 


20 


) -8V3846 8 


22 


) 4\/250a 6 


24 


) -\/512n 6 


26 


) V64u 5 t> 3 


28 


) \/l000a 4 6 5 


30 


) -\/l89x 3 y 6 


32 


) -4 3 a/56x 2 2/ 8 


34 


) 8V- 750x|/ 


36 


) 3Vl35a:2/ 3 


38 


— 6\/80m 4 p 7 g 4 


40 


) -6V405a 5 6 8 c 


42 


) -6V324x 7 y2 7 



294 



8.3 

Radicals - Adding Radicals 

Objective: Add like radicals by first simplifying each radical. 

Adding and subtracting radicals is very similar to adding and subtracting with 
variables. Consider the following example. 

Example 388. 

5x + 3x — 2x Combine like terms 
6x Our Solution 



5v 11 + 3v 11 — 2vH Combine like terms 
6vTT Our Solution 



Notice that when we combined the terms with VI 1 it was just like combining 
terms with x. When adding and subtracting with radicals we can combine like 
radicals just as like terms. We add and subtract the coefficients in front of the 



295 



radical, and the radical stays the same. This is shown in the following example. 

Example 389. 

7\/6 + 4V3-9\/3 + \/6 Combine like radicals 7 V6 + V6 and 4 ^3 - 8 \/3 
8 \/6 — 5 \/3 Our Solution 



We cannot simplify this expression any more as the radicals do not match. Often 
problems we solve have no like radicals, however, if we simplify the radicals first 
we may find we do in fact have like radicals. 



Example 390. 



5 v 45 + 6 v 18 — 2 v 98 + v 20 Simplify radicals, find perfect square factors 

5\/9 T 5 + 6\/9 T 2 - 2V49 • 2 + v / 4 T 5 Take roots where possible 

5 • 3\/5 + 6 • 3\/2 - 2 • 7\/2 + 2y/E Multiply coefficients 

15\/5 + 18\/2-14\/2 + 2\/5 Combine like terms 

17\/5 + 4\/2 Our Solution 



World View Note: The Arab writers of the 16th century used the symbol sim- 
ilar to the greater than symbol with a dot underneath for radicals. 

This exact process can be used to add and subtract radicals with higher indices 



Example 391. 



4 \/54 — 9 \/l6 + 5 \/9 Simplify each radical, finding perfect cube factors 

4 V27-2 - 9 'W^ + 5 V9 Take roots where possible 

4-3^-9-2^ + 5^9 Multiply coefficients 

12 V2 - 18 \/2 + 5 V9 Combine like terms 12 y/2 - 18 y/2 

- 6 V2 + 5 V9 Our Solution 



296 



8.3 Practice - Adding Radicals 



Simiplify 

I) 2\/5+2\/5 + 2\/5 

3) - 3\/2 + 3\/5 + 3\/5 
5) - 2\/6 - 2\/6 - \/6 
7) 3\/6 + 3\/5 + 2\/5 
9) 2\/2 - 3\/l8 - \/2 

II) -3\/6-\/l2+3\/3 
13) 3\/2 + 2\/8-3\/l8 
15) 3\/l8 - \/2 - 3\/2 

17) - 3\/6 - 3\/6 - \/3 + 3\/6 

19) - 2\/l8 - 3\/8 - \/20 + 2 v / 20 

21) - 2\/24 - 2\/6 + 2\/6 + 2 v / 20 

23) 3\/24 - 3\/27 + 2\/6 + 2\/8 

25) - 2Vl6 + 2\/l6 + 2n/2 

27) 2a/243 - 2V243 - VS 

29) 3a/2 - 2a/2 - V243 

31) - V324 + 3a/324 - 3a/4 

33) 2V2 + 2V3 + 3V64 - i/S 

35) - 3\/6 - V64 + 2 Vl92 - 2\/64 

37) 2^/160 - 2^192 - Vl60 - V^W) 

39) - V256 - 2Vi - 3V320 - 2VI28 



54-3\/6 + 3\/27 
\/5 - VE - 2\/54 



2) - 3\/6 - 3\/3 - 2\/3 
4) - 2\/6 - V3 - 3\/6 

6) -3\/3+2\/3-2\/3 
8) - y/5 + 2\/3 - 2\/3 

10 
12 
14 
16 
18 
20 
22 
24 
26 
28 
30 
32 
34 
36 
38 



2^20 + 2\/20-\/3 

- 3\/27 + 2\/3 - Vl2 

-2\/2-\/2 + 3\/8 + 3\/6 
-3\/l8-\/8 + 2\/8 + 2\/8 

- 3\/8 - \/5 - 3\/6 + 2\/l8 
2\/6 - \/54 - 3\/27 - \/3 
3\/l35 - \/8l - Vl35 



- 3Vi + 3V324 + 2V64 

2V6 + 2V4 + 3V6 

- 2V243 - V96 + 2V96 



2V48 - 3a/405 - 3\/48 - Vl62 

- 3V3 - 3V768 + 2\/384 + 3 V5 

- 2V256 - 2V256 - 3\/2 - V640 



297 



8.4 

Radicals - Multiply and Divide Radicals 

Objective: Multiply and divide radicals using the product and quotient 
rules of radicals. 

Multiplying radicals is very simple if the index on all the radicals match. The 
prodcut rule of radicals which we have already been using can be generalized as 
follows: 

Product Rule of Radicals: a r \/b • c r \/d = ac r \/bd 

Another way of stating this rule is we are allowed to multiply the factors outside 
the radical and we are allowed to multiply the factors inside the radicals, as long 
as the index matches. This is shown in the following example. 

Example 392. 

— 5V14 • 4\/6 Multiply outside and inside the radical 

— 20 v84 Simplify the radical, divisible by 4 

— 20\/4 • 21 Take the square root where possible 

- 20 • 2\/21 Multiply coefficients 

- 40\/2T Our Solution 



The same process works with higher roots 
Example 393. 



o i o i 

2vl8-6vl5 Multiply outside and inside the radical 

1 2 V270 S implify the radical , divisible by 2 7 

12 \/27 • 10 Take cube root where possible 

3 



12 • 3 V 10 Multiply coefficients 
36 vXO Our Solution 



When multiplying with radicals we can still use the distributive property or FOIL 
just as we could with variables. 

Example 394. 



7v6(3v 10 — 5v 15) Distribute, following rules for multiplying radicals 

21 v60 — 35 v90 Simplify each radical, finding perfect square factors 

21\/4 • 15 — 35\/9 • 10 Take square root where possible 

21 • 2V15 - 35 • 3VT0 Multiply coefficients 

42v / 15 - 105\/T0 Our Solution 



Example 395. 

( v5 — 2v3) (4vi~0 + 6 v6) FOIL, following rules for multiplying radicals 



298 



4v 50 + 6 v 30 — 8v 30 — 12 v 18 Simplify radicals, find perfect square factors 

4V25-2 + 6\/30 - 8\/30 - 12\/9 : 2 Take square root where possible 

4 • 5\/2 + 6\/30 - 8 V30 - 12 • 3\/2 Multiply coefficients 

20\/2 + 6\/30-8\/30-36\/2 Combine like terms 

-16\/2-2\/30 Our Solution 



World View Note: Clay tablets have been discovered revealing much about 
Babylonian mathematics dating back from 1800 to 1600 BC. In one of the tables 
there is an approximation of v2 accurate to five decimal places (1.41421) 

Example 396. 

(2 V5 — 3 v6) (7v2 — 8 v7) FOIL, following rules for multiplying radicals 

14\/l0 — 16\/35 — 21\/l2 — 24\/42 Simplify radicals, find perfect square factors 

14\/X0 - 16\/35 - 21V1T3 - 24\/42 Take square root where possible 

Uy/10 - 16\/35 - 21 • 2\/3 - 24\/42 Multiply coefficient 

14\/X0 - 16\/35 - 42 V3 - 24 v / 42 Our Solution 



As we are multiplying we always look at our final solution to check if all the radi- 
cals are simplified and all like radicals or like terms have been combined. 

Division with radicals is very similar to multiplication, if we think about division 
as reducing fractions, we can reduce the coefficients outside the radicals and 
reduce the values inside the radicals to get our final solution. 



Quotient Rule of Radicals: 



cVd c V d 



Example 397. 



— Reduce -— and — =— by dividing by 5 and 2 respectively 

20 V2 20 y/2 

Simplify radical, 54 is divisible by 27 



3\/54 



3\/27-2 
4 

3-3V2 
4 

9^2 
4 



Take the cube root of 27 



Multiply coefficients 



Our Solution 



There is one catch to dividing with radicals, it is considered bad practice to have 
a radical in the denominator of our final answer. If there is a radical in the 
denominator we will rationalize it, or clear out any radicals in the denominator. 

299 



We do this by multiplying the numerator and denominator by the same thing. 
The problems we will consider here will all have a monomial in the denominator. 
The way we clear a monomial radical in the denominator is to focus on the index. 
The index tells us how many of each factor we will need to clear the radical. For 
example, if the index is 4, we will need 4 of each factor to clear the radical. This 
is shown in the following examples. 

Example 398. 



y/6 



Index is 2, we need two fives in denominator, need 1 more 



V5\ 



Multiply numerator and denominator by \/5 



30 



Our Solution 



Example 399. 



3VTT 

V2 



Index is 4, we need four twos in denominator, need 3 more 



3Vn(W 



V2 4 v¥ 



Multiply numerator and denominator by v 2 3 



31/88 



Our Solution 



Example 400. 

4^2 



The 25 can be written as 5 . This will help us keep the numbers small 

7 V25 



4\/2 
7VP 



Index is 3, we need three fives in denominator, need 1 more 



4\/2 / V5 



3/q / : ., 

— — I ir-= Multiply numerator and denominator by v5 
7 vP V Vo / 



4VlO 

7-5 



Multiply out denominator 



4VlO 
35 



Our Solution 



300 



The previous example could have been solved by multiplying numerator and 

denominator by v25 2 . However, this would have made the numbers very large 
and we would have needed to reduce our soultion at the end. This is why re- 
writing the radical as v 5 2 and multiplying by just Vo was the better way to sim- 
plify. 

We will also always want to reduce our fractions (inside and out of the radical) 
before we rationalize. 



Example 401. 



6V14 





12y/22 




V7 




2VTE 


v/7. 


<Vn\ 


2VTP 


v ynJ 




v/77 




2-11 




^77 



Reduce coefficients and inside radical 



Index is 2, need two elevens, need 1 more 



Multiply numerator and denominator by v 11 



22 



Multiply denominator 



Our Solution 



The same process can be used to rationalize fractions with variables. 
Example 402. 



18 i/6x 3 y 4 z 
8 \Zl0xy & z 3 



Reduce coefficients and inside radical 



i/oZs 



9 v 3x Index is 4. We need four of everything to rationalize, 



4 \/5y 2 z 3 t hr ee mor e fives , two mor e y 's and one more z . 



9 V3X 2 {\/&ifz\ r 
— , Multiply numerator and denominator by v 5 

4 4 v / 5y 2 ^VV5V^ / 



3 y 2 z 



9 \/375x 2 y 2 z 
4- byz 



Multiply denominator 



9 \/375x 2 y 2 z 
20yz 



Our Solution 



301 



8.4 Practice - Multiply and Divide Radicals 



Multiply or Divide and Simplify. 

I) 3v / 5--4v / 16 
3) Vl2m-Vl5m 
5) Vi^ 1 • V2^ 
7) v / 6(v / 2 + 2) 

9) -5v / 15(3v / 3 + 2) 

II) 5v / 10(5n + v / 2) 

13) (2 + 2v / 2)(-3 + v / 2) 
15) (v / 5-5)(2 v / 5-l) 
17) (v / 2a + 2v / 3a)(3v / 2a- 
19) (-5-4v / 3)(-3-4v / 3) 



5a) 



12 



^ v 


5^/I00 


23) 


4VT25 


25) 


^/I0 
7^ 


27) 


2y/4 

3^3 


29) 


5X 2 


4 v / 3x 3 y 3 


31) 


V3p 


33) 


3\/T0 
5\/27 


35) 


3 ^ 
4^4 


*\7) 


5V&^ 



5V10-V15 



4) Vbr* ■ - 5v / 10r 
6) 3^4^ • VlOa 5 



/ 8r 2 



10) 
12) 
14) 
16) 
18) 
20) 
22) 

24) 

26) 

28) 

30) 

32) 
34) 

36) 
38) 



10(^/5 + \/2) 
5v / 15(3v / 3+2) 
Vl5(V5 - 3Vlv) 

(-2 + \/3)(-5 + 2v / 3) 
(2v / 3 + V / 5)(5v / 3 + 2 V / 4) 
( - 2 v 7 ^ + 5\/5) (\/5p + \/5p) 
(5v / 2-l)(-v / 2>^ + 5) 



15 

2VI 

•/L2 

3^5 

4^3 
%/l5 



5^/3* J/4 
•/[On 



15 

64 

V2 



2V64 

4 
V64m 



302 



8.5 

Radicals - Rationalize Denominators 

Objective: Rationalize the denominators of radical expressions. 

It is considered bad practice to have a radical in the denominator of a fraction. 
When this happens we multiply the numerator and denominator by the same 
thing in order to clear the radical. In the lesson on dividing radicals we talked 
about how this was done with monomials. Here we will look at how this is done 
with binomials. 

If the binomial is in the numerator the process to rationalize the denominator is 
essentially the same as with monomials. The only difference is we will have to dis- 
tribute in the numerator. 



Example 403. 

n/3-9 



2\/6 



Want to clear v6 in denominator, multiply by v6 



——?-{ — — We will distribute the \/6 through the numerator 



303 



18 — 9^/6 

Simplify radicals in numerator, multiply out denominator 



2-6 

\ZiT2-9v / 6 
12 

3\/2-9^6 
12 

\/2--3vj 
4 



Take square root where possible 
Reduce by dividing each term by 3 
Our Solution 



It is important to remember that when reducing the fraction we cannot reduce 
with just the 3 and 12 or just the 9 and 12. When we have addition or subtrac- 
tion in the numerator or denominator we must divide all terms by the same 
number. 

The problem can often be made easier if we first simplify any radicals in the 
problem. 



2\/20:c 5 — \l\2x 2 

Simplify radicals by finding perfect squares 



\/l8 



:c 



2\/4-5:r 3 -\/4-3:r 2 Q . ... . ,. ., . , . 
-=^ Simplify roots, divide exponents by 2. 

V 9 • 2x 

— Multiply coefficients 



Multiplying numerator and denominator by v 2x 



3V2x 
4x 2 ^5x - 2x\/3 



3V22: 

== -I . Distribute through numerator 

Z^2x \V2x J 

4x 2 v 10a: 2 — 2x v6a? Simplify roots in numerator, 

3 • 2x multiply coefficients in denominator 

4x 3 VT0 - 2x\/6a; 



6x 



Reduce, dividing each term by 2x 



304 



2xV 10 - V6x 

Our Solution 

3x 

As we are rationalizing it will always be important to constantly check our 
problem to see if it can be simplified more. We ask ourselves, can the fraction be 
reduced? Can the radicals be simplified? These steps may happen several times 
on our way to the solution. 

If the binomial occurs in the denominator we will have to use a different strategy 
to clear the radical. Consider —= — , if we were to multiply the denominator by 

v 3 — 5 _ ' 

Vo we would have to distribute it and we would end up with 3 — 5v3- We have 
not cleared the radical, only moved it to another part of the denominator. So our 
current method will not work. Instead we will use what is called a conjugate. A 
conjugate is made up of the same terms, with the opposite sign in the middle. 
So for our example with v3 — 5 in the denominator, the conjugate would be v3 + 
5. The advantage of a conjugate is when we multiply them together we have 
(v3 — 5)(v3 + 5), which is a sum and a difference. We know when we multiply 
these we get a difference of squares. Squaring v3 and 5, with subtraction in the 
middle gives the product 3 — 25 = — 22. Our answer when multiplying conjugates 
will no longer have a square root. This is exactly what we want. 

Example 404. 

2 

Multiply numerator and denominator by conjugate 

Distribute numerator, difference of squares in denominator 

Simplify denoinator 

Reduce by dividing all terms by — 2 

Our Solution 



\/3-5 

\/3 + 
\/3-5^\/3 + 5 

2ygj-10 
3-25 

2yg+10 

-22 

-\/3-5 

11 



In the previous example, we could have reduced by dividng by 2, giving the solu- 
tion , both answers are correct. 

Example 405. 



\/5 + \/3 



Multiply by conjugate, v5 — v3 



305 



Distribute numerator, denominator is difference of squares 



5 + \/3 V y/5-VsJ 

75-\/45 
5-3 



Simplify radicals in numerator, subtract in denominator 



Take square roots where possible 



2 

5\/3-3\/5 
2 

Example 406. 

2\/3x 



Our Solution 



4- V5a^ 



2\/3x /4 + V5P 



4 - V 5x 3 \4 + V5x 3 J 
8^3x + 2VTox 3 



Multiply by conjugate, 4 + v 5x 3 



Distribute numerator, denominator is difference of squares 



16 — 5x 3 

Sx + 2x 2 v / 15 
16 — 5x 3 



Simplify radicals where possible 



Our Solution 



The same process can be used when there is a binomial in the numerator and 
denominator. We just need to remember to FOIL out the numerator. 



Example 407. 

3-\/5 
2 ->/3 

3-Vb f 2 + V3 \ 
2-\/3\2 + \/3 J 

6 + 3\/3 - 2\/5 - \/l5 



4-3 

6 + 3\/3-2\/5-\/l5 



Multiply by conjugate, 2 + \/3 



FOIL in numerator, denominator is difference of squares 



Simplify denominator 



Divide each term by 1 



6 + 3\/3-2\/5-\/i~5 Our Solution 



306 



Example 408. 

2^ -3\/7 

5\/6+4\/2 



Multiply by the conjugate, 5\/6 — 4\/2 



2\/5 - 3\/7 ( 5yj - 4yg \ FOIL numerator, 

5.^/5 _|_ 4^/2 I 5\/6 — 4\/2 / denominator is difference of squares 



10V30-8V10-15\/42 + 12Vl4 .. u . . . , 

Multiply in denominator 



Subtract in denominator 



Our Solution 
118 





25-6-16-2 


lO-v/30- 


- 8^10 - 15\/42 + 12 v / 14 




150-32 


10^30- 


- 8\/T0 - 15\/42 + 12v / 14 



The same process is used when we have variables 
Example 409. 



3xy2x + v 4:X^ i 

Multiply by the conjugate, 5x + v 3x 



3£\/2~r + v4a? / 5x + \/3~r \ FOIL in numerator, 



5^ _ y/3x \ 5x + \/3x / denominator is difference of squares 

15x 2 \/2x + ZxVOx 1 + 5x\/ia^ + Vl2a; 4 



Simplify radicals 
Divide each term by x 



Our Solution 
25a; — 3 



World View Note: During the 5th century BC in India, Aryabhata published a 
treatise on astronomy. His work included a method for finding the square root of 
numbers that have many digits. 



307 



25x 2 — 3x 




\hx 2 \Flx + 3x 2 \/6 + 10x 2 v 


r x+2x 2 \/3 


25x 2 — 3x 




lbx\/2x + 3x\/6 + 10xi 


/x+2xV3 



8.5 Practice - Rationalize Denominators 



Simplify. 

-.X 4 + 2^3 2^ - 4 + ^3 

' s/% ' 4^9 

ox 4 + 2^3 ,s 2^-2 

' 5\/4 ' 2-/16 

5) ^ 6) ^ 

7 4^13 7 4^17 

^2-3^3 Q \ V5-V2 



7) 



\/3 ' 3^6 



27) 
29) 



^a + ^5 

2 + x/6 
2 + ^3 

a — Vo 
a+Vb 



3V 



3^2-2^3 

33) ""V 

a\fb — by/a 

2-V5 
-3 + V5 



35) 



9) 5 10) - 

' 3^ + \/2 ' \/3+4y5 

11) — ?-= 12) 



5 + V2 ' 2^-^ 

13 ) db 14) 7^ 

15 ) i^W 16 ^ 2^5+2^3 

17) - 4 18) -=X-= 

> 4-4^2 7 4^-^ 

19) -^-= 20) ^^ 

7 l + \/2 ^ 73-1 

91 \ V^4-2 , 2 + ^10 

j 77-^ ^ 72>W 



23) ^~° 24) vJL " vT 

25) 



Vb-y/a ' VT4 + V7 

fl + Vab 



26) 


a + vab 


^ + ^ 


28) 


2^5 + V3 


1-V3 


30) 


a — b 


Va + Vb 


32) 


ab 


ayo — by/a 


34) 


4V2 + 3 


3^2 + V3 


36) 


-l + %/5 


2^5+5^2 



308 



37) 



5^ + V^ 
5 + 5^2 



38) 



V3 + V2 



309 



8.6 



Radicals - Rational Exponents 



Objective: Convert between radical notation and exponential notation 
and simplify expressions with rational exponents using the properties 
of exponents. 

When we simplify radicals with exponents, we divide the exponent by the index. 
Another way to write division is with a fraction bar. This idea is how we will 
define rational exponents. 

n 

Definition of Rational Exponents: a™ = ( r \fa) n 

The denominator of a rational exponent becomes the index on our radical, like- 
wise the index on the radical becomes the denominator of the exponent. We can 
use this property to change any radical expression into an exponential expression. 

Example 410. 



- 3 

(Vx) 3 = x 5 


(V3^) 5 =(3x)^ 


1 _£ 
-= = a 7 

( 7 v^) 3 


1 _ 2 

(T1l\ 3 


(X/x-yf [Xy) 



Index is denominator 

Negative exponents from reciprocals 



We can also change any rational exponent into a radical expression by using the 
denominator as the index. 

Example 411. 



5_ 

a* 



(W 



x 



(tyx) 



(2m n) 



(\/2mn) 2 



(xy)' 



(%/xyf 



Index is denominator 

Negative exponent means reciprocals 



World View Note: Nicole Oresme, a Mathematician born in Normandy was the 

first to use rational exponents. He used the notation -»9 P to represent 9 3 . How- 
ever his notation went largely unnoticed. 

The ability to change between exponential expressions and radical expressions 
allows us to evaluate problems we had no means of evaluating before by changing 
to a radical. 

Example 412. 

4 

27 3 Change to radical, denominator is index, negative means reciprocal 



(V27) 



Evaluate radical 



310 



(3) 4 
81 



Evaluate exponent 



Our solution 



The largest advantage of being able to change a radical expression into an expo- 
nential expression is we are now allowed to use all our exponent properties to sim- 
plify. The following table reviews all of our exponent properties. 

Properties of Exponents 



a m a n = a m+n 


(ab) m = a m b m 


a~ m = — 
a m 


a m 

— = a m - n 

a n 


/a\ m a m 
\b) ~~W 


1 -a m 
a~ m 

/ a\~ m b m 
\~b) ~~a™ 


(a m ) n = a mn 


a°=l 



When adding and subtracting with fractions we need to be sure to have a 
common denominator. When multiplying we only need to multiply the numera- 
tors together and denominators together. The following examples show several 
different problems, using different properties to simplify the rational exponents. 



Example 413. 



2 111 

a3 fe^ a e bi 

4 5 12 

a ¥ &To a e })W 

5 7 



Need common denominator on a's (6) and b's (10) 
Add exponents on a's and b's 
Our Solution 



Example 414. 



1 2 
•3 7/5 



x 3 y 5 j Multiply — by each exponent 



Example 415. 



i — 

X iyw Our Solution 



x 2 y 3 -2x 2 y 6 



7 

x?y G 



In numerator, need common denominator to add exponents 



311 



4 4 15 

X 2 I/®- 2X 2 y~6 

7 

x*y° 

1 £ 
2x2 ye 

7 
X 2 



Add exponents in numerator, in denominator, y° = 1 



Subtract exponents on x , reduce exponent on y 



2x X 2/ 2 Negative exponent moves down to denominator 



2y 



;r 



Our Solution 



Example 416. 

i 

I 2 \ ~2 

25x 3 y 5 | Using order of operations, simplify inside parenthesis first 



9x 5 y 



4 £ 

2" 



Need common denominators before we can subtract exponents 



_5_ 4 \ " 
■15 yW \ 



Subtract exponents, be careful of the negative: 
4 / 15 \ 4 15 19 



25x 

12 15~ I 

g x T5y-wJ 10 \^ 10 )~ 10 ' 10 ~ 10 



25x 15 1/ 10 



The negative exponent will flip the fraction 



9 V 1 

— — ^ I The exponent — goes on each factor 



25x 15 y 10 
9 2 



1 7 19 
25 2 x~3o y2o 



T_ 

3x 30 

IF 

5y™ 



Evaluate 9 2 and 25 2 and move negative exponent 



Our Solution 



It is important to remember that as we simplify with rational exponents we are 
using the exact same properties we used when simplifying integer exponents. The 
only difference is we need to follow our rules for fractions as well. It may be worth 
reviewing your notes on exponent properties to be sure your comfortable with 
using the properties. 



312 



8.6 Practice - Rational Exponents 

Write each expression in radical form. 

3 3 

1) ms 2) (lOr) - * 

3) (7z)* 4) (66)"^ 

Write each expression in exponential form. 

5) 1 6) ^ 



7 ) Wky 8 ) ^ 

Evaluate. 

9) 8^ 10) 16* 

11) £ 12) 100"^ 

Simplify. Your answer should contain only positive exponents. 



13) yx 3 ■ xy 2 


15) (ah^)~ l 


»)£ 


3 

19) uv-u- (v 2 ) 3 


21) (arV)V 


3 7 
0Q ^ alb- 1 -M 
Z6 > 3/.-1 



25) 



5_ 

32/ 4 



14) 4w 3 -f" 1 


16) (:r%- 2 ) 


i i 

is) 2x ~: v \ 

2x3y 4 


20) (x-xt/ 2 ) 


22) u~*v 2 -(u 2 )~ 2 


24) rr x 



I/- 1 - 2y 3 26) 

28) 




30) 

(s 3 j/ 3 • J/) -1 <J^J 



_ 4 1 

31 



33) m 34) 



x- iy 3 . ^2 


1 5 

ab3-2b 4 


1 2 

4a 2 b 3 


1 3 
(2/ 2 ) 2 


3 1 

x 2 2/ 2 


2/° 


3 1 
(x4j/-l)3 


1 4 
(sV) 3 


2 

y 4 ■ x~ 2 y 3 


f i/V a ^ 



5 _3 
(*V) 2 



313 



8.7 

Radicals - Radicals of Mixed Index 



Objective: Reduce the index on a radical and multiply or divide radi- 
cals of different index. 

Knowing that a radical has the same properties as exponents (written as a ratio) 
allows us to manipulate radicals in new ways. One thing we are allowed to do is 
reduce, not just the radicand, but the index as well. This is shown in the fol- 
lowing example. 

Example 417. 



yxy Rewrite as rational exponent 

i 
(x 6 y 2 )s Multiply exponents 

6 2 



x s y s Reduce each fraction 

1 i 

x J y J All exponents have denominator of 4, this is our new index 
yx 3 y Our Solution 

What we have done is reduced our index by dividing the index and all the expo- 
nents by the same number (2 in the previous example). If we notice a common 
factor in the index and all the exponnets on every factor we can reduce by 
dividing by that common factor. This is shown in the next example 

Example 418. 

v a 6 fr 9 c 15 Index and all exponents are divisible by 3 
Va 2 b 3 c 5 Our Solution 

We can use the same process when there are coefficients in the problem. We will 
first write the coefficient as an exponential expression so we can divide the 
exponet by the common factor as well. 



Example 419. 



V8m 6 n 3 Write 8 as 2 3 

v 2 3 m 6 n 3 Index and all exponents are divisible by 3 
v 2m 2 n Our Solution 



We can use a very similar idea to also multiply radicals where the index does not 
match. First we will consider an example using rational exponents, then identify 
the pattern we can use. 

314 



Example 420. 



Vab 2 va 2 b Rewrite as rational exponents 

i i 

(ab 2 ) 3 (a 2 bY Multiply exponents 



h 2 ) 3 (a 2 

12 2 1 



a 3 b 3 a 4 b 4 To have one radical need a common denominator, 12 

JL jl jl jl 
a i2 fri2 a i2 ^12 Write under a single radical with common index, 1 2 

v a 4 b 8 a 6 b 3 Add exponents 

vVW 1 Our Solution 

To combine the radicals we need a common index (just like the common denomi- 
nator). We will get a common index by multiplying each index and exponent by 
an integer that will allow us to build up to that desired index. This process is 
shown in the next example. 

Example 421. 

va 2 b 3 va 2 b Common index is 12. 

Multiply first index and exponents by 3, second by 2 
v a 6 b 9 a 4 b 2 Add exponents 
vVOfc 11 Our Solution 

Often after combining radicals of mixed index we will need to simplify the 
resulting radical. 

Example 422. 

\/x 3 y 4 yx 2 y Common index: 15. 

Multiply first index and exponents by 3, second by 5 
V x 9 y 12 x 10 y 5 Add exponents 

■\/x 19 y 17 Simplify by dividing exponents by index, remainder is left inside 
xy \/ x A y 2 Our Solution 

Just as with reducing the index, we will rewrite coefficients as exponential expres- 
sions. This will also allow us to use exponent properties to simplify. 

Example 423. 

\J^x 2 y \/8xy 3 Rewrite 4 as 2 2 and 8 as 2 3 

\/2 2 x 2 y i/2 3 xy 3 Common index: 12. 

Multiply first index and exponents by 4, second by 3 

a/2 8 :t 8 ?/ 4 2 9 £ 3 ?/ 9 Add exponents (even on the 2) 

Y 2 17 x u y 13 Simplify by dividing exponents by index, remainder is left inside 

2y 1 $2 5 x 11 y Simplify 2 5 

2 y Y / 32x 11 y Our Solution 



315 



If there is a binomial in the radical then we need to keep that binomial together 
through the entire problem. 



Example 424. 



y/3x(y + z) \/9x(y + z) 2 Rewrite 9 as 3 2 

y/3x(y + z) \/?> 2 x(y + z) 2 Common index: 6 . Multiply first group by 3 , second by 2 

\/3 3 x 3 (y + z) 3 3 4 x 2 (y + z) 4 Add exponents, keep (y + z) as binomial 

\/3 7 x 5 (y + z) 7 Simplify, dividing exponent by index, remainder inside 

3(j/ + z) \/?>x b (y + z) Our Solution 



World View Note: Originally the radical was just a check mark with the rest of 
the radical expression in parenthesis. In 1637 Rene Descartes was the first to put 
a line over the entire radical expression. 

The same process is used for dividing mixed index as with multiplying mixed 
index. The only difference is our final answer cannot have a radical over the 
denominator. 



Example 425. 



v^w 



\/x 7 y 2 z 



Common index is 24. Multiply first group by 4, second by 3 



2 4/ x 16 y 12 z 8 



x 21y6 z 3 



Subtract exponents 



\/ x 5 y G z b Negative exponent moves to denominator 



W v z 

\ — g- Cannot have denominator in radical, need 12x's , or 7 more 



ryD 1 \j rp Li) 



"■"' ' ' "■ l ' ' Multiply numerator and denominator by \l x 7 



V 7 ^ ■' ' 



'x^y"z° 



x 



Our Solution 



316 



8.7 Practice - Radicals of Mixed Index 



Reduce the 


following radicals. 




1) Vl6zV 


2) V9sV 


3) ^64xV* 6 
5)V£ 


4 / 25x 3 
' V 16s 5 


6) fx 9 y 12 z 6 


7) \fx 6 y 9 


8) ^64aV 


9) V*V* 2 


10) V25y 2 


li) Vs^ 3 y 6 


12) ^81zV 2 



Combine the following radicals. 



13 

15 
17 
19 
21 

23 
25 
27 
29 
31 
33 
35 

37 
39 

41 
43 

45 



5\/6 
Xy/7y 



VI 



xyx 



V~x 2 * 



x Vy/xy 

\fx~tf\[xh) 



\/a 2 bc 2 Va 2 b 3 c 



4 A~3 
ay a 



WW 3 
\fx^\J~x 2 y 



V9ab 3 VSa^b 



\/Zxy 2 z\j9x 3 yz 2 
y/27a 5 (b+i) 3 y/Sla(b+l) 



?/^2 



V*V 


\/ab 3 c 


\/a 2 b 3 c _1 


V(3x-1) 3 


V(3^-l) 3 


V(2x + l) a 

5 //r.. ,W 



14 
16 
18 
20 
22 
24 
26 
28 
30 
32 
34 
36 

38 
40 

42 
44 
46 



V7V5 

3/1/ V3^ 

V3^\/y + 4 
Va& V2a 2 6 2 
Vo¥ Vo 
y/x 2 yz 3 %Jx 2 yz 2 

3 / o 6 / S 

vr vr 



4 A~3 3 A~2 
5 / TT / r 

va o vao 



y/2x 3 y 3 y/Axy 2 
\/a% z c A Vab 2 c 

y/Sx(y + zf %jAx 2 (y + z) 2 



VJ 





V^V* 9 


^/xy~ 2 z 


V(2 + 5xf 


V(2 + 5x) 


V(5-3x) s 

■A 1 In a \9 



317 



8.8 

Radicals - Complex Numbers 

Objective: Add, subtract, multiply, rationalize, and simplify expres- 
sions using complex numbers. 

World View Note: When mathematics was first used, the primary purpose was 
for counting. Thus they did not originally use negatives, zero, fractions or irra- 
tional numbers. However, the ancient Egyptians quickly developed the need for "a 
part" and so they made up a new type of number, the ratio or fraction. The 
Ancient Greeks did not believe in irrational numbers (people were killed for 
believing otherwise). The Mayans of Central America later made up the number 
zero when they found use for it as a placeholder. Ancient Chinese Mathematicians 
made up negative numbers when they found use for them. 

In mathematics, when the current number system does not provide the tools to 
solve the problems the culture is working with, we tend to make up new ways for 
dealing with the problem that can solve the problem. Throughout history this has 
been the case with the need for a number that is nothing (0), smaller than zero 
(negatives), between integers (fractions), and between fractions (irrational num- 
bers). This is also the case for the square roots of negative numbers. To work 
with the square root of negative numbers mathematicians have defined what are 
called imaginary and complex numbers. 

Definition of Imaginary Numbers: i 2 = — 1 (thus i = V '— 1) 

Examples of imaginary numbers include 3i, — 6i, -i and 3iv5- A complex 
number is one that contains both a real and imaginary part, such as 2 + 5i. 

With this definition, the square root of a negative number is no longer undefined. 
We now are allowed to do basic operations with the square root of negatives. 
First we will consider exponents on imaginary numbers. We will do this by 
manipulating our definition of i 2 = — 1. If we multiply both sides of the definition 
by i, the equation becomes i 3 = — i. Then if we multiply both sides of the equa- 
tion again by i, the equation becomes i A = — i 2 = — ( — 1) = 1, or simply z 4 = 1. 
Multiplying again by i gives i 5 = i. One more time gives i 6 = i 2 = — 1. And if this 
pattern continues we see a cycle forming, the exponents on i change we cycle 
through simplified answers of i, — 1, — i, 1. As there are 4 different possible 
answers in this cycle, if we divide the exponent by 4 and consider the remainder, 
we can simplify any exponent on i by learning just the following four values: 

Cyclic Property of Powers of i 

i = i 
i 2 = -l 
i 3 = — i 



318 



Example 426. 



Example 427. 



z 35 Divide exponent by 4 

\R3 Use remainder as exponent on i 

i 3 Simplify 

— i Our Solution 



i 124 Divide exponent by 4 

31 RO Use remainder as exponent on i 

i° Simplify 

1 Our Solution 

When performing operations (add, subtract, multilpy, divide) we can handle i just 
like we handle any other variable. This means when adding and subtracting com- 
plex numbers we simply add or combine like terms. 

Example 428. 

(2 + hi) + (4 — li) Combine like terms 2 + 4 and hi — li 
6 — 2i Our Solution 

It is important to notice what operation we are doing. Students often see the 
parenthesis and think that means FOIL. We only use FOIL to multiply. This 
problem is an addition problem so we simply add the terms, or combine like 
terms. 

For subtraction problems the idea is the same, we need to remember to first dis- 
tribute the negative onto all the terms in the parentheses. 

Example 429. 

(4 — 8i) — (3 — hi) Distribute the negative 

4 — 8i — 3 + hi Combine like terms 4 — 3 and — 8i + hi 
1 — 3i Our Solution 

Addition and subtraction can be combined into one problem. 

Example 430. 

(hi) — (3 + 8i) + ( — 4 + li) Distribute the negative 

hi — 3 — 8i — 4 + li Combine like terms hi — 8i + li and — 3 — 4 
— 7 + 4i Our Solution 

Multiplying with complex numbers is the same as multiplying with variables with 
one exception, we will want to simplify our final answer so there are no exponents 
on i. 



319 



Example 431. 



(3i) (7i) Multilpy coefficients and i's 

21i 2 Simplify i 2 = - 1 

21( — 1) Multiply 

— 21 Our Solution 



Example 432. 



5i{3i - 7) 

15i 2 — 35z 

15(-l)-35i 

- 15 - 35* 



Distribute 
Simplify i 2 = — 1 
Multiply 
Our Solution 



Example 433. 



(2-4z)(3 + 5z) 

6 + l0i-l2i-20i 2 

10z-12i-20(-l) 

6 + 10z-12z + 20 

26 -2i 



1 



FOIL 

Simplify i 2 ■■ 
Multiply 

Combine like terms 6 
Our Solution 



20andl0z-12i 



Example 434. 



(3i)(6i)(2-3i) 

18* 2 (2-3z) 

36i 2 - 54i 3 

36(-l)-54(-i) 

- 36 + 54i 



Multiply first two monomials 

Distribute 

Simplify i 2 = — 1 and i 3 = — i 

Multiply 

Our Solution 



Remember when squaring a binomial we either have to FOIL or use our shortcut 
to square the first, twice the product and square the last. The next example uses 
the shortcut 



Example 435. 



(5if 



(4-5i) 2 

4 2 = 16 

2(4)(-5i) = -40i 

25z 2 = 25(-l) = -25 

16-40z-25 

- 9 - 40z 



Use perfect square shortcut 

Square the first 

Twice the product 

Square the last, simplify i 2 = 

Combine like terms 

Our Solution 



320 



Dividing with complex numbers also has one thing we need to be careful of. If i is 
\J — 1, and it is in the denominator of a fraction, then we have a radical in the 
denominator! This means we will want to rationalize our denominator so there 
are no i's. This is done the same way we rationalized denominators with square 
roots. 

Example 436. 



7 + 3i 




-5i 


7 + 3i, 


'i\ 


-hi ' 


kO 


li + 3i 2 


— 


5i 2 


7i + 3(- 


-1) 


-5(- 


1) 


1% 


-3 



Just a monomial in denominator, multiply by i 



Distribute i in numerator 



Simplify ft 



Multiply 



Our Solution 



The solution for these problems can be written several different ways, for example 
or — — h -i, The author has elected to use the solution as written, but it is 



5 5 ' 5 

important to express your answer in the form your instructor prefers. 



Example 437. 



2-6i 

4 + 8i 



Binomial in denominator, multiply by conjugate, 4 — 8i 



- — —I - — — I FOIL in numerator, denominator is a difference of squares 
4 + 8* V 4-8i ' 



8 - 16i - 24i + 48f 





16 - 64i 2 


8- 


-16z-24i + 48(-l) 




16-64(-l) 




8 - 16i - 24i - 48 




16 + 64 




- 40 - 40i 




80 




-l-i 



Simplify i 2 



Multiply 



Combine like terms 8 — 48 and — 16i — 24i and 16 + 64 



Reduce, divide each term by 40 



Our Solution 



321 



Using i we can simplify radicals with negatives under the root. We will use the 
product rule and simplify the negative as a factor of negative one. This is shown 
in the following examples. 



Example 438. 



\J — 16 Consider the negative as a factor of — 1 
\/— 1 • 16 Take each root , square root of — 1 is % 
4i Our Solution 



Example 439. 



\J — 24 Find perfect square factors, including — 1 
\J — 1-4-6 Square root of — 1 is i , square root of 4 is 2 
2z\/6 Our Solution 

When simplifying complex radicals it is important that we take the — 1 out of the 
radical (as an i) before we combine radicals. 

Example 440. 

\J — 6y/— 3 Simplify the negatives, bringing i out of radicals 

(i v6) (iv3) Multiply i by i is i 2 = — 1 , also multiply radicals 

— \/l8 Simplify the radical 
— \/9 • 2 Take square root of 9 

— 3\/2 Our Solution 

If there are fractions, we need to make sure to reduce each term by the same 
number. This is shown in the following example. 



Example 441. 



-15- 


-V-200 


20 
V-200 


V- 


-1-100-2 

10i\/2 

>-10zv / 2 





20 
3-2n/2 




4 



Simplify the radical first 

Find perfect square factors, including — 1 
Take square root of — 1 and 100 
Put this back into the expression 

All the factors are divisible by 5 
Our Solution 



By using % = \J — 1 we will be able to simplify and solve problems that we could 
not simplify and solve before. This will be explored in more detail in a later sec- 
tion. 



322 



8.8 Practice - Complex Numbers 



Simplify. 

1) 3-(-8 + 4i) 
3) (7i) - (3 - 2i) 
5) (-6i)-(3 + 7i) 
7) (3 — 3z) + ( — 7 — 8i) 



9)(< 

11 
13) 

15) 
17) 
19) 
21) 

23) 
25) 

27) 
29) 



- (2 + 3i) — 6 

6i)(-8i) 

-5i)(8i) 

-lif 
6 + 5i) 2 

-7-4i)(-8 + 6i) 
-4 + 5i)(2-7i) 
-8-6i)(-4 + 2i) 
1 + 50(2 + 

-9 + 5i 



31) 

33) 
35) 
37) 
39) 
41) 



- 10 -9i 
%i 

-3-6i 

4i 

10 -i 



1/ 



- 10 + i 

8 
7-Gi 

7 
10-7? 

5i 
-6-i 



2) (3<) - (7t) 




4) 5 + (-6-6i) 




6) ( - 8<) - (7t) - 


-(5-30 


8)(-4-0 + (l 


-50 


10) (5-4i) + (8 


-40 


12) (30(-80 




14) (80(-40 




16) (-0(70(4- 


30 


18) (80(-20(- 


2-80 


20) (30(-30(4 


-40 


22) -8(4-80- 


-2(-2-6i 


24) (-60(3-2i 


)- (70(40 


26) (-2 + 0(3- 


50 


28) " 3 t 2i 

7 — 3i 




30) ~\ +2t 

7 3t 




32) ~ 5 a +9i 

7 9t 




34) H 




36) T 9 \. 

7 1 — 5« 




38 ) 4 + 6* 




40) 9 

7 -8-6i 




42) 8i 
7 6-7?; 





323 



43) x/= 


81 


45) V- 


10V-2 


47 3W 




=^27 


49) 8 "^ 


/ :r T6 
4 


51) z 73 




53) i 48 




55) z 62 




57) z 154 





V- 


45 




V- 


■12</= 


-2 


-4- 


-V=8 






-4 




6 + -. 


/^32 





52) i 251 
54) i 68 
56) i 181 
58) i 51 



324 



Chapter 9 : Quadratics 

9.1 Solving with Radicals 326 

9.2 Solving with Exponents 332 

9.3 Complete the Square 337 

9.4 Quadratic Formula 343 

9.5 Build Quadratics From Roots 348 

9.6 Quadratic in Form 352 

9 . 7 Application: Rectangles 357 

9.8 Application: Teamwork 364 

9.9 Simultaneous Products 370 

9.10 Application: Revenue and Distance 373 

9.11 Graphs of Quadratics 380 



325 



9.1 



Quadratics - Solving with Radicals 



Objective: Solve equations with radicals and check for extraneous solu- 
tions. 

Here we look at equations that have roots in the problem. As you might expect, 
to clear a root we can raise both sides to an exponent. So to clear a square root 
we can rise both sides to the second power. To clear a cubed root we can raise 
both sides to a third power. There is one catch to solving a problem with roots in 
it, sometimes we end up with solutions that do not actually work in the equation. 
This will only happen if the index on the root is even, and it will not happen all 
the time. So for these problems it will be required that we check our answer in 
the original problem. If a value does not work it is called an extraneous solution 
and not included in the final solution. 

When solving a radical problem with an even index: check answers! 



Example 442. 



V7X + 2 


= 4 


7a: + 2 = 
-2 


= 4 2 
= 16 

-2 



Even index! We will have to check answers 

Square both sides, simplify exponents 

Solve 

Subtract 2 from both sides 



7a; = 14 


Divide both sides by 7 


T" T~ 




x = 2 


Need to check answer in original problem 


^7(2) + 2 = 4 


Multiply 


^14 + 2 = 4 


Add 


v / 16 = 4 


Square root 


4 = 4 


True! It works! 


2 = 2 


Our Solution 


Example 443. 




Vx - 1 = - 4 


Odd index, we don't need to check answ 


(Va;-l) 3 = (-4) 3 


Cube both sides, simplify exponents 


x - 1 = - 64 


Solve 



326 



+ 1 +1 Add 1 to both sides 
x = — 63 Our Solution 



Example 


i 444. 






V3x + 6 = 


-3 




(V3x + 6) = (- 


3) 4 




3a; + 6 = 


= 81 




-6 


-6 




Sx = 


= 75 




~3~ 


~3~ 




X- 


= 25 




V3(25)+6 = 


-3 




4 V75 + 6 = 


-3 




V8l = 


-3 




3 = 


-3 




No Solul 


ion 



Even index! We will have to check answers 

Rise both sides to fourth power 

Solve 

Subtract 6 from both sides 

Divide both sides by 3 

Need to check answer in original problem 

Multiply 

Add 

Take root 

False, extraneous solution 

Our Solution 



If the radical is not alone on one side of the equation we will have to solve for the 
radical before we raise it to an exponent 



Example 445. 



x + \/4x + 1 = 5 Even index! We will have to check solutions 
jc — x Isolate radical by subtracting x from both sides 

Square both sides 

Evaluate exponents, recal (a — b) 2 = a 2 — 2ab + b 2 

Re — order terms 

Make equation equal zero 

Subtract 4x and 1 from both sides 

Factor 

Set each factor equal to zero 

Solve each equation 



\/4x + 1 = 5 - X 


W4x + l) 2 


= (5 -a:) 2 


4x + l = 25 


-lOx + a; 2 


4x + 1 = x 2 - 


-10a; + 25 


-4x-l 


-4x - 1 


= x 2 - 


- 14a; + 24 


= (aj- 


12)(x-2) 


ar -12=0 OI 


■ a:-2 = 


+ 12 + 12 


+ 2 + 2 



x = 12 or x = 2 Need to check answers in original problem 
(12) + ^4(12) + 1=5 Check x = 5 first 



327 



12 + \/48TT = 5 Add 

12 + \/49 = 5 Take root 

12 + 7 = 5 Add 

19 = 5 False, extraneous root 

(2) + v / 4( 2 ) + 1=5 Checkx = 2 

2 + x/8+T = 5 Add 

2 + \/9 = 5 Take root 

2 + 3 = 5 Add 

5 = 5 True! It works 

x = 2 Our Solution 



The above example illustrates that as we solve we could end up with an x 2 term 
or a quadratic. In this case we remember to set the equation to zero and solve by 
factoring. We will have to check both solutions if the index in the problem was 
even. Sometimes both values work, sometimes only one, and sometimes neither 
works. 

World View Note: The babylonians were the first known culture to solve 
quadratics in radicals - as early as 2000 BC! 

If there is more than one square root in a problem we will clear the roots one at a 
time. This means we must first isolate one of them before we square both sides. 



Example 446. 

\/3x — 8 — y/x = Even index! We will have to check answers 

+ y/x + y/x Isolate first root by adding y/x to both sides 

\/3x — 8 = y/x Square both sides 

(\/3x — 8) 2 = (\/x) 2 Evaluate exponents 

3x — 8 = x Solve 

-3x —3x Subtract 3x from both sides 

— 8 = — 2x Divide both sides by — 2 
=~2 ^2 

4 = x Need to check answer in original 

^3(4) -8 - \fl = Multiply 

V12-8 - \/l = Subtract 

y/i - y/A = Take roots 



328 



■2 = Subtract 
= True! It works 
x = 4 Our Solution 



When there is more than one square root in the problem, after isolating one root 
and squaring both sides we may still have a root remaining in the problem. In 
this case we will again isolate the term with the second root and square both 
sides. When isolating, we will isolate the term with the square root. This means 
the square root can be multiplied by a number after isolating. 



Example 447. 

\/2x + 1 — \fx = 1 Even index! We will have to check answers 

+ \fx + i/x Isolate first root by adding \px to both sides 

\/2x + 1 = \fx + 1 Square both sides 

(a/2x + 1) 2 = (y/x + l) 2 Evaluate exponents, recall (a + b) 2 = a 2 + 2o b + b 2 

2x + 1 = x + 2^fx + 1 Isolate the term with the root 

- x — 1 — x — 1 Subtract x and 1 from both sides 

x = 2^Jx Square both sides 

(x) 2 = (2y/x) 2 Evaluate exponents 

x 2 = 4x Make equation equal zero 

— 4x — 4x Subtract x from both sides 

x 2 — Ax = Factor 

x(x — 4) = Set each factor equal to zero 

x = or x — 4 = Solve 

+ 4 + 4 Add 4 to both sides of second equation 

x = or x = 4 Need to check answers in original 

x /2(0) + l- v / (0) = l Checkx = 0first 

y/l - V0 = 1 Take roots 

1 — = 1 Subtract 

1 = 1 True! It works 

^2(4) + 1- v / (4) = 1 Checkx = 4 

^/8+T - \fl = 1 Add 

y/9 - Vl = 1 Take roots 

3 — 2 = 1 Subtract 

1 = 1 True! It works 

329 



x = or 4 Our Solution 



Example 448. 



y/3x + 9-y/x + 4 = -l 


Even index! We will have to check answers 


+ \/x + 4 + yjx + 4 


Isolate the first root by adding \J x + 4 


V3a; + 9 = Vx + 4 - 1 


Square both sides 


(V3x + 9)^ = (Vx + 4-l) 2 


Evaluate exponents 


3x + 9 = a; + 4- 2\/x + 4 + 1 


Combine like terms 


3x + 9 = x + 5-2\/x + 4 


Isolate the term with radical 


— x — 5 — x — 5 


Subtract x and 5 from both sides 


2a; + 4 = - 2y/x + 4 


Square both sides 


(2x + 4) 2 =(-2Vx + 4) 2 


Evaluate exponents 


4x 2 + 16x + 16 = 4(x + 4) 


Distribute 


4x 2 + 16a; + 16 = 4a: + 16 


Make equation equal zero 


-4x-16-4x-16 


Subtract 4x and 16 from both sides 


4x 2 + 12x = 


Factor 


4x(x + 3) = 


Set each factor equal to zero 


4x = or x + 3 = 


Solve 


T T -3-3 




x = or x = — 3 


Check solutions in original 


V / 3(0) + 9- v / (0)+4 = -l 


Check x = first 


x /9_74 = _l 


Take roots 


3-2=-l 


Subtract 


1 = -1 


False, extraneous solution 


'3(-3)+9- v / (-3)+4 = -l 


Check x = — 3 


V-9 + 9- v / (-3)+4 = -l 


Add 


\/0-\/l = -l 


Take roots 


0- 1= - 1 


Subtract 


-1 = -1 


True! It works 


x = — 3 


Our Solution 



330 



9.1 Practice - Solving with Radicals 

Solve. 

1) V2:c + 3-3 = 2) y/bx + 1 -4 = 



3) VQx-5 -x = 4) v 7 ^ 1 ^ - y/x = 2 

5) 3 + x = V6x + 13 6)x-l = \/f^x 

7) V3-3X - 1 = 2x 8) v / 2x + 2 = 3 + V2x - 1 

9) \/4^T5 - v^+4 = 2 10) V3x + 4 - v^+2 = 2 

11) v / 2^+4- v / ^T3 = l 12) v / 7x + 2- v / 3x + 6 = 6 

13) \/2x + 6 - \/^+4 = 1 14) y/Ax-Z - VZx + 1 = 1 

15) V6-2X - v / 2a 7 +3 = 3 16) V2 - 3a; - \/3x + 7 = 3 



331 



9.2 

Quadratics - Solving with Exponents 

Objective: Solve equations with exponents using the odd root property 
and the even root property. 

Another type of equation we can solve is one with exponents. As you might 
expect we can clear exponents by using roots. This is done with very few unex- 
pected results when the exponent is odd. We solve these problems very straight 
forward using the odd root property 

Odd Root Property: if a n = b, then a = \/o when n is odd 
Example 449. 



x 5 = 32 


Use odd root property 


W> = V32 


Simplify roots 


x = 2 


Our Solution 



However, when the exponent is even we will have two results from taking an even 
root of both sides. One will be positive and one will be negative. This is because 
both 3 2 = 9 and ( — 3) 2 = 9. so when solving x 2 = 9 we will have two solutions, one 
positive and one negative: x = 3 and — 3 



Even Root Property: if a n = b, then a = ± vb when n is even 
Example 450. 



re 4 = 16 Use even root property ( ± 



332 



= ± V16 Simplify roots 
x = ± 2 Our Solution 



World View Note: In 1545, French Mathematicain Gerolamo Cardano pub- 
lished his book The Great Art, or the Rules of Algebra which included the solu- 
tion of an equation with a fourth power, but it was considered absurd by many to 
take a quantity to the fourth power because there are only three dimensions! 



Example 451. 

(2x + 4) 2 = 36 Use even root property ( ± ) 

y/{2x + 4) 2 = ± \/36 Simplify roots 

2x + 4 = ± 6 To avoid sign errors we need two equations 

2x + 4 = 6 or 2x + 4 = — 6 One equation for + , one equation for — 

— 4 — 4 —4 —4 Subtract 4 from both sides 



2x = 2 or 2x = - 


-10 


Divide both si< 


~2 ^2 ~2 


~2 




x = l or x = 


-5 


Our Solutions 



In the previous example we needed two equations to simplify because when we 
took the root, our solutions were two rational numbers, 6 and — 6. If the roots did 
not simplify to rational numbers we can keep the ± in the equation. 



Example 452. 

(6x — 9) 2 = 45 Use even root property ( ± ) 

\J {§x — 9) 2 = ± v45 Simplify roots 

6x — 9 = ± 3 v5 Use one equation because root did not simplify to rational 

+ 9+9 Add 9 to both sides 



6x = 9±3v5 Divide both sides by 6 



I) 6 

~6 

3±\/5 



9 ± 3\/5 
x = Simplify, divide each term by 3 



Our Solution 



333 



When solving with exponents, it is important to first isolate the part with the 
exponent before taking any roots. 



Example 453. 








(x + 4) 3 -6 = 119 


Isolate part with exponent 




+ 6 +6 






(x + 4) 3 = 125 


Use odd root property 




V(^ + 4) 3 = \/l25 


Simplify roots 




£ + 4 = 5 


Solve 




-4-4 


Subtract 4 from both sides 




x= 1 


Our Solution 



Example 454. 



(6a; +1) 



6 = 10 
-6-6 



6.x 





(6x + l) 2 


= 4 




y/(6 X +l) 2 = ± 


v/4 




6x + 1 = 


±2 


• + 1 


= 2 or 6x + 1 = 


-2 


-1 


-1 -1 


-1 



6x : 
~6~ 

x- 



1 or 6x : 



6 

1 
= 6 



6 



or x ■ 



Isolate part with exponent 

Subtract 6 from both sides 

Use even root property ( ± ) 

Simplify roots 

To avoid sign errors, we need two equations 

Solve each equation 

Subtract 1 from both sides 

Divide both sides by 6 

Our Solution 



When our exponents are a fraction we will need to first convert the fractional 

m 

exponent into a radical expression to solve. Recall that a~ = (y/a) m . Once we 
have done this we can clear the exponent using either the even ( ± ) or odd root 
property. Then we can clear the radical by raising both sides to an exponent 
(remember to check answers if the index is even). 



Example 455. 



(Ax 

(\fl~x~ 



m 



x - 



1)5=9 

T) 2 = 9 
I) 2 = ±\/9 



Rewrite as a radical expression 

Clear exponent first with even root property ( ± 

Simplify roots 



334 



%/Ax + 1 = ± 3 


Clear radical by raising both sides to 5tl 


(V4x + 1) 5 = (±3) 5 


Simplify exponents 


Ax + 1 = ± 243 


Solve, need 2 equations! 


4x + 1 = 243 or 4x + 1 = - 243 




-1 -1 -1 -1 


Subtract 1 from both sides 


Ax = 242 or 4x = - 244 


Divide both sides by 4 


XXX ~4~ 




121 fil 

x= , 61 


Our Solution 


Example 456. 




3 

(3x — 2) 1 = 64 Rewrite as radical expression 


(y/3x — 2) 3 = 64 Clear exponent first with odd root property 


\l(i/3x-2)' i =V64: Simplify roots 


\/3x — 2 = 4 Even Index! Check answers. 


(y/3x — 2 ) 4 = 4 4 Raise both sides to 4th power 


3x - 2 = 256 Solve 


+ 2 +2 Add 2 to both sides 


3a; = 258 Divide both sides by 3 



3 3 

x = 86 Need to check answer in radical form of problem 

(V3(86)-2) 3 = 64 Multiply 

(V258-2) 3 = 64 Subtract 

(V256) 3 = 64 Evaluate root 

4 3 = 64 Evaluate exponent 

64 = 64 True! It works 

x = 86 Our Solution 



With rational exponents it is very helpful to convert to radical form to be able to 
see if we need a ± because we used the even root property, or to see if we need 
to check our answer because there was an even root in the problem. When 
checking we will usually want to check in the radical form as it will be easier to 
evaluate. 



335 



9.2 Practice - Solving with Exponents 



Solve. 

I) a: 2 = 75 

3) a: 2 + 5 = 13 

5) Sx 2 + 1 = 73 

7) (a; + 2) 5 = -243 

9) (2x + 5) 3 -6 = 21 

II) (£-1)^ = 16 
13) (2-^ = 27 
15) (2x-3)^ = 4 
17)(a: + i)-t = 4 
19) (x-l)~* = 32 
21) (3a; -2)^= 16 
23) (4x + 2)^ = -8 



2) x- 


5 = -8 


4) 4a; 3 - 2 = 106 


6) (x-4) 2 = 49 


8)(E 


>a; + l) 4 = 16 


10) 


;2x + l) 2 + 3 = 21 


12) ( 


£-1)^=8 


14) ( 


2x + 3)a = 16 


16) ( 


x + 3)~* = 4 


18) ( 


x-l)~* = Z2 


20) ( 


a; + 3)2 = -8 


22) 


;2x + 3)^ = 27 


24) ( 


3 -2a;)* =-81 



336 



9.3 



Quadratics - Complete the Square 



Objective: Solve quadratic equations by completing the square. 

When solving quadratic equations in the past we have used factoring to solve for 
our variable. This is exactly what is done in the next example. 



Example 457. 



x 2 + 5a: + 6 = Factor 

(x + 3) (x + 2) = Set each factor equal to zero 

x + 3 = or :r + 2 = Solve each equation 

-3-3 -2-2 

x = — 3 or x = — 2 Our Solutions 



However, the problem with factoring is all equations cannot be factored. Consider 
the following equation: x 2 — 2x — 7 = 0. The equation cannot be factored, however 
there are two solutions to this equation, 1 + 2V2 and 1 — 2v2. To find these two 
solutions we will use a method known as completing the square. When completing 
the square we will change the quadratic into a perfect square which can easily be 
solved with the square root property. The next example reviews the square root 
property. 



Example 458. 





(x + 


5) 2 - 


= 18 


y/(x + 5) 2 -- 


= ±- 


v/18 




x + 5 = 


= ±; 


iy/2 




-5 


-5 





X- 



Square root of both sides 
Simplify each radical 
Subtract 5 from both sides 



5±3\/2 Our Solution 



337 



To complete the square, or make our problem into the form of the previous 
example, we will be searching for the third term in a trinomial. If a quadratic is 
of the form x 2 + bx + c, and a perfect square, the third term, c, can be easily 

found by the formula ( - • b ) . This is shown in the following examples, where we 

find the number that completes the square and then factor the perfect square. 



Example 459. 



ix- 



x 



■ b I and our b = 8 



= 4 2 = 16 The third term to complete the square is 16 

- 8a; + 16 Our equation as a perfect square, factor 
(x + 4) 2 Our Solution 



Example 460. 



x 2 — 7x + c c = I — • b 1 and our 6=7 



1 \ 2 /7\ 2 49 49 

— • 7 I = I — ) = — The third term to complete the square is — — 



49 
x 2 — 1 lx + — - Our equation as a perfect square, factor 



■> — — I Our Solution 



Example 461. 



5 



1 



x + —x + c c = I — ■ • b \ and our 6 = 8 



— • — ) = ( — I = tttt The third term to complete the square is -— 
2 3 / \ 6 / 36 36 



338 



x 2 + — x + —— Our equation as a perfect square, factor 
3 36 



x - 



Our Solution 



The process in the previous examples, combined with the even root property, is 
used to solve quadratic equations by completing the square. The following five 
steps describe the process used to complete the square, along with an example to 
demonstrate each step. 



Problem 


3x 2 +18x-6 = 


1. Separate constant term from variables 


+ 6 + 6 
3x 2 + 18x =6 


2. Divide each term by a 


3 2 . 18 6 
3 "^ 3 X ~~ 3 

x 2 + 6x =2 


3. Find value to complete the square: ( - • b J 


(i-6) 2 = 3 2 = 9 


4. Add to both sides of equation 


x 2 + 6x =2 
+ 9 +9 
x 2 + 6x + 9 = ll 


5. Factor 


(x + 3) 2 = ll 


Solve by even root property 


y/(x + 3) 2 = ±Vll 
x + 3 = ±\/iT 
-3 -3 

x = -3±v / ll 



World View Note: The Chinese in 200 BC were the first known culture group 
to use a method similar to completing the square, but their method was only used 
to calculate positive roots. 

The advantage of this method is it can be used to solve any quadratic equation. 
The following examples show how completing the square can give us rational solu- 
tions, irrational solutions, and even complex solutions. 



Example 462. 



2:r 2 + 20a; + 48 = Separate constant term from varaibles 



339 



-48-48 Subtract 24 

2x 2 + 2Cte = - 48 Divide by a or 2 

~2~ ~T ~2~ 

(\ x2 

x 2 + lOrc = — 24 Find number to complete the square: f — • b 

1 \ 2 

— •101 =5 2 = 25 Add 25 to both sides of the equation 

x 2 +10a: =-24 
+ 25 +25 

x 2 + lOx + 25 = 1 Factor 

(x + 5) 2 = 1 Solve with even root property 

yj{x + 5) 2 = ± \/T Simplify roots 

a; + 5 = ± 1 Subtract 5 from both sides 

-5-5 

x = — 5 ± 1 Evaluate 

x = — 4 or — 6 Our Solution 



Example 463. 

x 2 — 3x — 2 = Separate constant from variables 
+ 2 + 2 Add 2 to both sides 



x 2 — Sx =2 No a , find number to complete the square I — • b 
1 A 2 (3\ 2 9 ,,,9 



2 



•3) = — =— Add — to both sides, 
2/12/4 4 



2/4\ 9__8 9_17 
l v 4/ + 4~4 + 4~ 4 



Need common denominator (4) on right 



2 o 9 8 9 17 ^ 
x— 3x + — = — + — = —- 1 actor 
4 4 4 4 



3 \ 2 17 

a; — — I = — — Solve using the even root property 



x — — ) = ± \ — Simplify roots 



3 ± V 17 3 

x — — = — - — Add — to both sides, 



340 



3 3 

2 + 2 



we already have a common denominator 



3 ± v 17 
x = Our Solution 



Example 464. 



3x 2 = 2x — 7 Separate the constant from the variables 

— 2x — 2x Subtract 2x from both sides 

3x 2 — 2x = — 7 Divide each term by a or 3 

T" ~3~ ~3~ 

2 7 (\ x2 

x 2 — — x = — — Find the number to complete the square I — • b 



1 2\ 2 /l\ 2 1 



2 3/ \ 3 / 9 



Add to both sides, 



7/3 \ 1 -21 1 -20 

+ — = — - — get common denominator on right 



313/9 3 9 9 



2 2 1 20 _ 

x — —x + — = — — 1 actor 
3 3 9 



1 \ 2 20 



x — — j = — — Solve using the even root property 

Of \j 



20 




Simplify roots 



1 ±2z^5 . . . 1 . ^ ., 

x = Add — to both sides, 

3 3 3 



1 1 
x = Our Solution 



Already have common denominator 



As several of the examples have shown, when solving by completing the square we 
will often need to use fractions and be comfortable finding common denominators 
and adding fractions together. Once we get comfortable solving by completing the 
square and using the five steps, any quadratic equation can be easily solved. 



341 



9.3 Practice - Complete the Square 

Find the value that completes the square and then rewrite as a perfect 
square. 

1) x 2 - 30x + 2) a 2 - 24a + 

3) m 2 -36m + 4) x 2 

5) x 2 — 15x+ _ 6) r 2 - 



34a; 
i 



7)y 2 -y + __ S) P 2 

Solve each equation by completing the square. 

9) x 2 - 16x + 55 = 



-r + 
9 ^ 

17p + __ 



11 


) v 2 -8v+45 = 


13 


) 6x 2 + 12x + 63 = 


15 


) hk 2 - lOJfe + 48 = 


17 


) x 2 + 10a;-57 = 4 


19 


) n 2 -16n + 67 = 4 


21 


) 2x 2 + Ax + 38 = - 6 


23 


) 86 2 + 166-37 = 5 


25 


) x 2 = -10x-29 


27 


) n 2 = -21 + 10n 


29 


) 3k 2 + 9 = 6k 


31 


) 2x 2 + 63 = 8x 


33 


p 2 — 8p = — 55 


35 


7n 2 — n + 7 = 7n + 6n 2 


37 


) 13fe 2 + 156 + 44 = -5 + 7 


39 


) 5x 2 + 5x = — 31 — 5x 


41 


) w 2 + 5i; + 28 = 


43 


) 7x 2 - 6x + 40 = 


45 


) k 2 -7k + 50 = 3 


47 


) 5x 2 + 8a; - 40 = 8 


49 


m 2 = — 15 + 9m 


51 


) 8r 2 + 10r = -55 


53 


) 5n 2 - 8n + 60 = - 3n + 6 


55 


) -2x 2 + 3x-5 = -4a; 2 



36 



4n 2 



10 
12 
14 
16 
18 
20 
22 
24 
26 
28 
30 
32 
34 
36 
38 
40 
42 
44 
46 
48 
50 
52 
54 
56 



n 2 -8n-12 = 
b 2 + 26 + 43 = 
3x 2 - 6x + 47 = 
8a 2 + 16a -1 = 
p 2 - 16p - 52 = 
m 2 — 8m — 3 = 6 
6r 2 +12r-24=-6 
6n 2 -12n-14 = 4 
v 2 = Uv + 36 
a 2 — 56 = — 10a 
5x 2 = -26 + 10x 
5n 2 = -10n + 15 
x 2 + 8x + 15 = 8 
n 2 + 4n = 12 
-3r 2 + 12r + 49 = - 
8n 2 + 16n = 64 
6 2 + 76 - 33 = 
4x 2 + 4x + 25 = 
a 2 - 5a + 25 = 3 
2j9 2 -p + 56 = -8 
n 2 — n = — 41 
3x 2 -lla; = -18 
46 2 -156 + 56 = 36 2 
10u 2 -15w = 27 + 4w 2 



6r 2 



6v 



342 



9.4 

Quadratics - Quadratic Formula 

Objective: Solve quadratic equations by using the quadratic formula. 

The general from of a quadratic is ax 2 + bx + c = 0. We will now solve this for- 
mula for x by completing the square 

Example 465. 

ax 2 + bc + c = Separate constant from variables 
- c — c Subtract c from both sides 
ax 2 + bx = — c Divide each term by a 

~a~ ~a~ ~a~ 

b — c 
x 2 -\ — x = Find the number that completes the square 



a a 

2 ' a J ~ I 2a J _ 4a 2 



4a \ 6 2 4ac 6 2 -4ac 



4a 2 a\Aa J 4a 2 4a 2 4a 2 



Add to both sides, 



Get common denominator on right 



2 b b 2 b 2 4ac_6 2 -4ac 
a 4a 2 4a 2 4a 2 4a 2 



b V b 2 -4ac 



x -\ = = — Solve using the even root property 

2a / 4a 2 



b \ , /5 2 -4ac 



x + -— ) = ± \ — — -pi — Simplify roots 
2a J V 4a 2 



x + -— = Subtract -— from both sides 

2a 2a 2a 



-b± Vb 2 -4ac „ a . ,. 

x = Our Solution 

2a 



This solution is a very important one to us. As we solved a general equation by 
completing the square, we can use this formula to solve any quadratic equation. 
Once we identify what a, 6, and c are in the quadratic, we can substitute those 

343 



— b i \/b 2 — 4ac 

values into x = 5 anc ^ we wm get our two solutions. This formula is 

known as the quadratic fromula 

Quadratic Formula: if ax z + bx + c = then cc = 



2 ; ,.... , „ n.u^.. -&±\/& 2 -4a C 



2a 

World View Note: Indian mathematician Brahmagupta gave the first explicit 
formula for solving quadratics in 628. However, at that time mathematics was not 
done with variables and symbols, so the formula he gave was, "To the absolute 
number multiplied by four times the square, add the square of the middle term; 
the square root of the same, less the middle term, being divided by twice the 
square is the value." This would translate to 

— ^— as the solution to the equation ax 2 + bx = c. 

We can use the quadratic formula to solve any quadratic, this is shown in the fol- 
lowing examples. 



Example 466. 



x 2 + 3x + 2 = a=l,b=3,c=2, use quadratic formula 

-3± V / 3 2 -4(l)(2) , , , 

x = — 7-t Evaluate exponent and multiplication 

2(1) 

3±V9 



x = Evaluate subtraction under root 

2 



x = Evaluate root 

-3±1 _ , 

x = Evaluate ± to get two answers 

— 2 —4 

x = —— or — — Simplify fractions 

x = — 1 or — 2 Our Solution 



As we are solving using the quadratic formula, it is important to remember the 
equation must fist be equal to zero. 

Example 467. 

25x 2 = 30x + ll First set equal to zero 
— 30a; — 11 — 30x — 11 Subtract 30a: and 11 from both sides 



25x 2 — 30x — 11 = a = 25, b= — 30, c= — 11, use quadratic formula 
30 ± v / (-30) 2 -4(25)(-ll) 



2(25) 



Evaluate exponent and multiplication 



344 



30±V900 + 1100 . 

x = Evaluate addition mside root 

50 

30±V2000 _. .., 

x = Simplify root 

50 

x = — Reduce fraction by dividing each term by 1 



50 
3_±2yg 
5 



Our Solution 



Example 468. 



3x 2 + 4x + 8 = 2x 2 + 6x — 5 First set equation equal to zero 
2x 2 — 6x + 5 — 2x 2 — 6x + 5 Subtract 2x 2 and 6x and add 5 



x — 2x+ 13 = a = l,b= — 2,c= 13, use quadratic formula 

2± v / (-2) 2 -4(l)(13) , , ,. , 

x = — t-t Evaluate exponent and multiplication 

2±\/^48 

re = Simplify root 

x = Reduce fraction by dividing each term by 2 

x = 1 ± 2iv3 Our Solution 



When we use the quadratic formula we don't necessarily get two unique answers. 
We can end up with only one solution if the square root simplifies to zero. 



Example 469. 



4x — \2x + 9 = a = 4, b = — 12, c = 9, use quadratic formula 

12± v / (-12) 2 -4(4)(9) „ , , ,. , 

x = ——t Evaluate exponents and multiplication 



2(4) _ 

12±\/0 



x = Evaluate subtraction inside root 



12 ±0 



Evaluate root 
Evaluate ± 



12 
x = — Reduce fraction 

3 

x = — Our Solution 
2 



345 



If a term is missing from the quadratic, we can still solve with the quadratic for- 
mula, we simply use zero for that term. The order is important, so if the term 
with x is missing, we have b = 0, if the constant term is missing, we have c= 0. 



Example 470. 



3x 2 + 7 = a = 3, b = (missing term), c = 7 
0± V / 2 -4(3)(7) 



2(3) 



x ■■ 



x ■■ 



±x/£g 

6 
±2n/2T 



6 
±n/2l 



X : 



Evaluate exponnets and multiplication, zeros not needed 
Simplify root 
Reduce, dividing by 2 
Our Solution 



We have covered three different methods to use to solve a quadratic: factoring, 
complete the square, and the quadratic formula. It is important to be familiar 
with all three as each has its advantage to solving quadratics. The following table 
walks through a suggested process to decide which method would be best to use 
for solving a problem. 



1. If it can easily factor, solve by factoring 



2. If a= 1 and b is even, complete the square 



3. Otherwise, solve by the quadratic formula 



x 



5x + 6 = 



(x-2)(x-3) = 

x = 2 or x = 3 



x 



2x = A 

.2 



•2 



1 2 =1 



2 

x 2 + 2x + l = h 
( x + l) 2 = 5 

x+i=±va 

x = -i±VE 



X 



X - 



x ■■ 



- 3x + 4 = 

3±7(-3)^-4(l)(4) 



2(1) 



3±iV7 



The above table is mearly a suggestion for deciding how to solve a quadtratic. 
Remember completing the square and quadratic formula will always work to solve 
any quadratic. Factoring only woks if the equation can be factored. 



346 



9.4 Practice - Quadratic Formula 



Solve each equation with the quadratic formula. 



1) 4a 2 + 6 = 


2) 


3A; 2 + 2 = 


3) 2x 2 - 8x - 2 = 


4) 


6n 2 - 1 = 


5) 2rn 2 - 3 = 


6) 5p 2 + 2p + 6 = 


7) 3r 2 - 2r - 1 = 


8) 


2a; 2 - 2x - 15 = 


9) An 2 - 36 = 


10 


) 3fe 2 + 6 = 


11) v 2 -4t>-5 = -8 


12 


) 2x 2 + 4x + 12 = 8 


13) 2a 2 + 3a + 14 = 6 


14 


) 6n 2 - 3n + 3 = - 4 


15) 3A; 2 + 3fc-4 = 7 


16 


) 4x 2 - 14 = - 2 


17) 7x 2 + 3x-16 = -2 


18 


) 4n 2 + 5n = 7 


19) 2p 2 + 6p-16 = 4 


20 


) m 2 + Am - 48 = - 3 


21) 3n 2 + 3n = -3 


22 


) 3b 2 - 3 = 86 


23) 2x 2 = -7x + 49 


24 


) 3 r 2 + 4 = _g r 


25) bx 2 = 7x + 7 


26 


) 6a 2 = - 5a + 13 


27) 8n 2 = -3n-8 


28 


) 6v 2 = 4 + 6v 


29) 2x 2 + 5x = -3 


30 


)x 2 = 8 


31) 4a 2 -64 = 


32 


) 2k 2 + 6k-16 = 2k 


33) 4p 2 + 5p - 36 = 3p 2 


34 


) 12a: 2 + x + 7 = 5x 2 + 5x 


35) -5n 2 -3n-52 = 2-7n 2 


36 


7m 2 — 6m + 6 = — m 


37) 7r 2 - 12 = - 3r 


38 


) 3x 2 - 3 = x 2 


39) 2n 2 -9 = 4 


40 


)Qb 2 = b 2 + 7-b 



347 



9.5 



Quadratics - Build Quadratics From Roots 



Objective: Find a quadratic equation that has given roots using reverse 
factoring and reverse completing the square. 

Up to this point we have found the solutions to quadratics by a method such as 
factoring or completing the square. Here we will take our solutions and work 
backwards to find what quadratic goes with the solutions. 

We will start with rational solutions. If we have rational solutions we can use fac- 
toring in reverse, we will set each solution equal to x and then make the equation 
equal to zero by adding or subtracting. Once we have done this our expressions 
will become the factors of the quadratic. 



Example 471. 




The solutions are 4 and — 2 


Set each solution equal to x 


x = 4 or x = — 2 


Make each equation equal zero 


-4-4 +2 +2 


Subtract 4 from first , add 2 to second 


£-4 = or £ + 2 = 


These expressions are the factors 


(x-4)(x + 2) = 


FOIL 


x 2 + 2x - Ax - 8 


Combine like terms 


x 2 -2x-8 = 


Our Solution 



If one or both of the solutions are fractions we will clear the fractions by multi- 
plying by the denominators. 



Example 472. 



2 3 
The solution are — and — 

3 4 

2 3 

x = — or x = — 

3 4 

3x = 2 or 4x = 3 



2-2 



3-3 



3x-2 = or 4x-3 
(3x - 2)(4x - 3) ■■ 
12x 2 - 9a; - 8x + 6 = 










Set each solution equal to x 

Clear fractions by multiplying by denominators 

Make each equation equal zero 

Subtract 2 from the first, subtract 3 from the second 

These expressions are the factors 

FOIL 

Combine like terms 



348 



12x 2 -17x + 6 = Our Solution 



If the solutions have radicals (or complex numbers) then we cannot use reverse 
factoring. In these cases we will use reverse completing the square. When there 
are radicals the solutions will always come in pairs, one with a plus, one with a 
minus, that can be combined into "one" solution using ± . We will then set this 
solution equal to x and square both sides. This will clear the radical from our 
problem. 



Example 473. 



The solutions are \/3 and — \/3 Write as "one" expression equal to x 
x = ± v3 Square both sides 



3 

3 



3 = 



Make equal to zero 
Subtract 3 from both sides 
Our Solution 



We may have to isolate the term with the square root (with plus or minus) by 
adding or subtracting. With these problems, remember to square a binomial we 
use the formula (a + b) 2 = a 2 + 2a 6 + b 2 



Example 474. 



The solutions are 2 



5\/2and2 + 5\/2 
x = 2±5\/2 
-2-2 

x-2 = ±5\/2 

x 2 - 4x + 4 = 25 • 2 

x 2 - 4x + 4 = 50 

-50-50 

2 



xr 



4 X - 46 = 



Write as "one" expression equal to x 
Isolate the square root term 
Subtract 2 from both sides 
Square both sides 

Make equal to zero 
Subtract 50 
Our Solution 



World View Note: Before the quadratic formula, before completing the square, 
before factoring, quadratics were solved geometrically by the Greeks as early as 
300 BC! In 1079 Omar Khayyam, a Persian mathematician solved cubic equations 
geometrically! 

If the solution is a fraction we will clear it just as before by multiplying by the 
denominator. 



349 



Example 475. 



The solutions are ■ 



4 



y/3 ,2 

and — 



y/3 



4 

2±\/3 



X 4 

4x = 2 ± y/3 

-2-2 



Ax - 2 = ± y/3 

16x 2 -16x + 4 = 3 

-3-3 

16x 2 -16x + l = 



Write as "one" expresion equal to x 

Clear fraction by multiplying by 4 

Isolate the square root term 
Subtract 2 from both sides 
Square both sides 
Make equal to zero 
Subtract 3 
Our Solution 



The process used for complex solutions is identical to the process used for radi- 
cals. 



Example 


476. 








The 


soluti 


ons are 


4 — 5i and 4 + 5z 








x = 


4±5i 








-4- 


4 








x-4 


= ±hi 






a; 2 


-8a; + 16 


= 25z 2 






x 2 - 


-8x + 16 = 
+ 25 


= -25 

+ 25 



ix- 



41 = 



Write as "one" expression equal to x 

Isolate the i term 

Subtract 4 from both sides 

Square both sides 

i 2 =-l 

Make equal to zero 

Add 25 to both sides 

Our Solution 



Example 477. 



The solutions are ■ 



5i .3 
— and 



5/ 



x = 

2x- 

-3- 



2 
3±5z 

2 
;3±5z 





2a; -3 = 


:±5Z 


4x 2 


-12a; + 9 


= 5i 2 


4x 2 - 


12a; + 9 = 


-25 




+ 25 


+ 25 



4x 2 - 12a; + 34 = 



Write as "one" expression equal to x 

Clear fraction by multiplying by denominator 

Isolate the i term 

Subtract 3 from both sides 

Square both sides 

i 2 =-l 

Make equal to zero 

Add 25 to both sides 

Our Solution 



350 



9.5 Practice - Build Quadratics from Roots 



From each problem, find a quadratic equation with those numbers as 
its solutions. 



1) 2, 5 
3) 20, 2 
5) 4, 4 
7)0,0 
9) -4,11 

n; 

13) 



15 

17 

19 
21 
23 

25 

27 

29 
31 
33 

35 

37 
39 



3 1 

4 ' 4 



1 1 
2 ' 3 



4 

1 5 
" 3 ' 6 

■6,i 



±5 



±V11 

i \/3 

4 

2±Ve 

l±3i 
6±iV3 

-l±\/6 
2 

6±t^ 



2) 


3, 6 


4) 


13, 1 


6) 


0, 9 


8) 


-2,-5 


10 


) 3, -1 


12 


. 5 5 
' 8 ' 7 


14 


1 2 
' 2 ' 3 


16 


)2 i 


18 


. 5 1 
' 3' _ 2 


20 


) --o 


22 


) ±1 


24 


) ±\/7 


26 


) ±2\/3 


28 


) ±lli 


30 


) ±5iV2 



32 
34 
36 

38 

40 



-3±V2 

-2±4z 

-9±i\/5 

2±5« 
3 

-2±i\/l5 



351 



9.6 



Quadratics - Quadratic in Form 



Objective: Solve equations that are quadratic in form by substitution 
to create a quadratic equation. 

We have seen three different ways to solve quadratics: factoring, completing the 
square, and the quadratic formula. A quadratic is any equation of the form = 
ax 2 + bx + c, however, we can use the skills learned to solve quadratics to solve 
problems with higher (or sometimes lower) powers if the equation is in what is 
called quadratic form. 

Quadratic Form: = ax m -\- bx n + c where m = 2n 

An equation is in quadratic form if one of the exponents on a variable is double 
the exponent on the same variable somewhere else in the equation. If this is the 
case we can create a new variable, set it equal to the variable with smallest expo- 
nent. When we substitute this into the equation we will have a quadratic equa- 
tion we can solve. 

World View Note: Arab mathematicians around the year 1000 were the first to 
use this method! 



Example 478. 








A 

x — 


13x 2 + 36 : 


= 






y = 


--x 2 






y 2 = 


--x A 




y 2 - 


■13y + 36 


= 




(y- 


9)(y-4). 


= 




y-9 = 


or y — 4 : 


= 




+ 9 + 9 


+ 4 + 4 




y 


= 9 or y 


= 4 




9 = 


x 2 or 4 = 


--x 2 


±V9-- 


= Vx 2 or 


±y/i = \ 


iH? 






X= ±3,: 


±2 



Quadratic form, one exponent, 4, double the other, 2 

New variable equal to the variable with smaller exponent 

Square both sides 

Substitute y for x 2 and y 2 for x 4 

Solve. We can solve this equation by factoring 

Set each factor equal to zero 

Solve each equation 

Solutions for y , need x. We will use y = x 2 equation 

Substitute values for y 

Solve using the even root property, simplify roots 

Our Solutions 



When we have higher powers of our variable, we could end up with many more 
solutions. The previous equation had four unique solutions. 



352 



Example 


479. 




a~ 2 


-a" 1 - 6 = 


= 




b = a 


-l 




b 2 = a 


-2 




b 2 -b-6 = 


= 


(b~ 


3)(6 + 2) = 


= 


fe-3 = 


or 6 + 2 = 


= 


+ 3 + 3 


-2- 


-2 


b = 


3 or b = - 


■2 


3 = a~ 1 


or — 2 = a 


-l 


3 _1 = a or 


(-2)" 1 = 


= a 




1 
a= 3'" 


1 

2 



Quadratic form, one exponent, — 2, is double the other, — 1 

Make a new variable equal to the variable with lowest exponent 

Square both sides 

Substitute b 2 for a~ 2 and b for a -1 

Solve. We will solve by factoring 

Set each factor equal to zero 

Solve each equation 

Solutions for b, still need a, substitute into b = a~ l 

Raise both sides to — 1 power 

Simplify negative exponents 

Our Solution 



Just as with regular quadratics, these problems will not always have rational solu- 
tions. We also can have irrational or complex solutions to our equations. 



Example 480. 



2x 4 + x 2 = 6 Make equation equal to zero 

— 6 — 6 Subtract 6 from both sides 

Quadratic form, one exponent, 4, double the other, 2 

New variable equal variable with smallest exponent 

Square both sides 

Solve. We will factor this equation 

Set each factor equal to zero 

2y — 3 = or y + 2 = Solve each equation 

3 + 3 - 

2y = 3 or y = — 2 

~2~ ~T 

3 
y = — or y = — 2 We have y , still need x. Substitute into y = x 2 



2x 4 + x 2 -6~- 


= 


y = 


x 2 


y 2 = 


4 
X 


2y 2 +y-6~- 


= 


(2y-3)(y + 2)-- 


= 


i-3 = or y + 2~- 


= 


+ 3 + 3 -2- 


-2 



2 

= x 
2 



3 

x 2 or —2 = x 2 Square root of each side 



/3 

± \/ — = Vx 2 or ± \J — 2 = v x 2 Simplify each root, rationalize denominator 

x = — - — , ± i\[2 Our Solution 



353 



When we create a new variable for our substitution, it won't always be equal to 
just another variable. We can make our substitution variable equal to an expres- 
sion as shown in the next example. 



Example 481. 



3(x — 7) 2 — 2(x — 7) + 5 = Quadratic form 

y = x — 7 Define new variable 

y 2 = (x — 7) 2 Square both sides 

3y 2 — 1y + 5 = Substitute values into original 

(3y-5)(y + l) = Factor 

3^ — 5 = or y + l = Set each factor equal to zero 

+ 5 + 5 —1 — 1 Solve each equation 

3y = 5 or y= — 1 

T~ ~3~ 

5 



5 

3 = 
21 

T 



t/ = — or y = — l We have y , we still need x. 
-x — 7 or — 1 =x — 7 Substitute into y = x — 7 

f7 +7 + 7 Add 7. Use common denominator as needed 

Our Solution 



26 « 



Example 482. 



(x 2 -&x) 2 = 7(x 2 -&x)-\2 

7(x 2 - 6x) + 12 - 7(x 2 - 6x) + 12 

(x 2 - 6x) 2 - 7(x 2 - 6x) + 12 = 

y = x 2 — 6x 

y 2 = (x 2 — Qx) 2 

y 2 -7y + \2 = 

(y-3)(y-4) = 

y — 3 = or y — 4 = 

+3+3 +4+4 



-or 



y = 3 or y = 4 

3 — 6a; 



6s or 4 = .r 

2 



1 



•6 



12: 



6x + 9 or 13 : 



:£ 



--3 2 -- 
6x- 



9 
9 



Make equation equal zero 

Move all terms to left 

Quadratic form 

Make new variable 

Square both sides 

Substitute into original equation 

Solve by factoring 

Set each factor equal to zero 

Solve each equation 

We have y , still need x. 

Solve each equation, complete the square 

Add 9 to both sides of each equation 

Factor 



354 



Use even root property 
Simplify roots 
Add 3 to both sides 

Our Solution 



The higher the exponent, the more solution we could have. This is illustrated in 
the following example, one with six solutions. 



12 = 


■ (x — 3) 2 or 13= (x - 


-3) 2 


= y/(x- 


3) 2 or ±y 


f W = V(x~- 


-3) 2 


±2\/3: 


= x — 3 or 


±Vl3 = a 


:-3 


+ 3 


+ 3 


+ 3 


+ 3 


X 


= 3±2\/3. 


3±\/l3 





Example 483. 




x 6 -9a; 3 + 8 


= 


y = 


--x 3 


y 2 = 


--x 6 


y 2 -9y + 8- 


= 


(y-l)(y-S)-. 


= 


y — 1 = or y — 8 


= 


+ 1 + 1 +8- 


+ 8 


y = 1 or y 


= 8 


x 3 = 1 or x 3 


= 8 


-1-1 -8 


-8 


x 3 — 1 = or x 3 — 8 


= 


(x-l)(x 2 + x + l) = i 





x — 1 = or x 2 + x + 1- 


= 


+ 1 + 1 




£= 1 




-1± V / 1 2 -4(1)(1) l±f 


v/3 


2(1) 2 




(x-2)(x 2 + 2a; + 4): 


= 


x - 2 = or x 2 + 2x + 4- 


= 


+ 2 + 2 




x = 2 




-2±V2 2 -4(1)(4) ^. 


y/3 


2(1) 


x-1,2, 1 ^, l±i 


Vs 



Quadratic form, one exponent, 6, double the other, 3 

New variable equal to variable with lowest exponent 

Square both sides 

Substitute y 2 for x 6 and y for x 3 

Solve. We will solve by factoring. 

Set each factor equal to zero 

Solve each equation 

Solutions for y, we need x. Substitute into y = x 3 

Set each equation equal to zero 

Factor each equation, difference of cubes 

First equation factored. Set each factor equal to zero 

First equation is easy to solve 

First solution 

Quadratic formula on second factor 

Factor the second difference of cubes 
Set each factor equal to zero. 
First equation is easy to solve 
Our fourth solution 

Quadratic formula on second factor 
Our final six solutions 



355 



9.6 Practice - Quadratic in Form 



Solve each of the following equations, 
complex roots. 

1) x 4 -5x 2 + 4 = 

3) m 4 - 7m 2 - 8 = 

5) a 4 - 50a 2 + 49 = 

7) x 4 - 25x 2 + 144 = 

9) m 4 - 20m 2 + 64 = 



11 
13 

15 
17 
19 
21 
23 
25 
27 
29 
31 
33 
35 
37 
39 

41 
43 

45 



216 = 192 3 



6^ 4 -^ 2 = 12 



2 1 

£3 - 35 = 2x3 
y- 6 + 7y- 3 = 8 
x 4 - 2x 2 - 3 = 
2a; 4 - 5x 2 + 2 = 
x 4 - 9x 2 + 8 = 
8x 6 - 9x 3 + 1 = 



x 



17a: 4 +16 = 



(y + b) 2 -4(y + b) = 21 
(y + 2) 2 -6(y + 2) = 16 
(x-3) 2 -2(x-3) = 35 
(r-l) 2 -8(r-l) = 20 
3(y+l) 2 -U(y + l)=5 
(Sx 2 -2x) 2 + 5 = 6(Sx 2 -2x) 

2(3x + l)^-5(3a: + 1)^ = 88 
(x 2 + 2a:) 2 -2(a; 2 + 2x) = 3 
(2x 2 -x) 2 -4(2x 2 -a:)+3 = 



Some equations will have 

2) y 4 -9y 2 + 20 = 
4) y 4 -29y 2 + 100 = 
6) 6 4 -106 2 + 9 = 
8) y 4 -A0y 2 + 144 = 



10 


) x 6 -35x 3 + 216 = 


12 


) y 4 _ 2y 2 = 24 


14 


) x" 2 -x- 1 -12 = 


16 


) 5y~ 2 -20 = 21y- 1 


18 


) x 4 -7x 2 + 12 = 


20 


) x 4 + 7x 2 + 10 = 


22 


) 2x 4 - x 2 - 3 = 


24 


) x 6 -10x 3 + 16 = 


26 


) 8x 6 + 7x 3 - 1 = 


28 


) (x-l) 2 -4(x-l) = 5 


30 


) (x + l) 2 + 6(x + l) + 9 = 


32 


) (m - l) 2 - 5(m - 1) = 14 


34 


) (a + l) 2 + 2(a-l) = 15 


36 


) 2(x-l) 2 -(x-l) = 3 


38 


) (x 2 -3) 2 -2(x 2 -3) = 3 


40 


) (x 2 + x + 3) 2 + 15 = 8(x 2 + x + 3) 


42 


) (x 2 + x) 2 -8(x 2 + x) + 12 = 


44 


) (2x 2 + 3x) 2 = 8(2x 2 + 3x) + 9 


46 


) (3x 2 - 4x) 2 = 3(3x 2 - 4x) + 4 



356 



9.7 

Quadratics - Rectangles 

Objective: Solve applications of quadratic equations using rectangles. 

An application of solving quadratic equations comes from the formula for the area 
of a rectangle. The area of a rectangle can be calculated by multiplying the width 
by the length. To solve problems with rectangles we will first draw a picture to 
represent the problem and use the picture to set up our equation. 

Example 484. 

The length of a rectangle is 3 more than the width. If the area is 40 square 
inches, what are the dimensions? 



40 



x We do not know the width, x. 



x + 3 Length is 4 more, or x + 4, and area is 40. 



357 



x(x + 3) = 40 Multiply length by width to get area 



x 



3x = 40 Distribute 



— 40 — 40 Make equation equal zero 

x 2 + 3a; - 40 = Factor 

(x — 5) (x + 8) = Set each factor equal to zero 

x — 5 = or £ + 8 = Solve each equation 
+5+5 -8-8 

x = 5 or x = — 8 Our x is a width, cannot be negative. 

(5) + 3 = 8 Length is x + 3, substitute 5 for x to find length 

5 in by 8 in Our Solution 



The above rectangle problem is very simple as there is only one rectangle 
involved. When we compare two rectangles, we may have to get a bit more cre- 
ative. 



Example 485. 

If each side of a square is increased by 6, the area is multiplied by 16. Find the 
side of the original square. 



x x Square has all sides the same length 



x 



Area is found by multiplying length by width 




x + 6 



x + 6 Each side is increased by 6 , 



Area is 16 times original area 



(x + 6) (x + 6) = 16x 2 Multiply length by width to get area 
x 2 + 12x + 36 = 16x 2 FOIL 
16x 2 — 16x 2 Make equation equal zero 



x ■■ 



— 15x 2 + 12x + 36 = Divide each term by — 1, changes the signs 
15x 2 — 12x — 36 = Solve using the quadratic formula 

12+ v / (-12) 2 -4(15)(-36) 



2(15) 



Evaluate 



x 



16 + V2304 



30 

x = — — — Can't have a negative solution, we will only add 
30 

x = — = 2 Our x is the original square 

OU 



358 



2 Our Solution 



Example 486. 

The length of a rectangle is 4 ft greater than the width. If each dimension is 
increased by 3, the new area will be 33 square feet larger. Find the dimensions of 
the original rectangle. 



We don't know width, x , length is 4 more, x + 4 
Area is found by multiplying length by width 






X 


+ 7 


(x + 


■3)(x + 7) = 


~-x(x + 4) +33 


x 2 


+ 10a; + 21 


= x 2 + 4a: + 33 


— x 2 




-x 2 




10a; + 21 = 4x + 33 




-4a; 


-Ax 

6x + 21 = 33 

-21-21 

6a: = 12 

"6" ~6~ 

x = 2 

(2) + 4 = 6 

2 ft by 6ft 



Increase each side by 3. 

width becomes a; + 3, length x + 4 + 3 = aH 

Area is 33 more than original , x (x + 4) + 33 
Set up equation, length times width is area 
Subtract x 2 from both sides 

Move variables to one side 
Subtract 4a; from each side 
Subtract 21 from both sides 

Divide both sides by 6 

x is the width of the original 

x + 4 is the length. Substitute 2 to find 

Our Solution 



7 



From one rectangle we can find two equations. Perimeter is found by adding all 
the sides of a polygon together. A rectangle has two widths and two lengths, both 
the same size. So we can use the equation P = 21 + 2w (twice the length plus 
twice the width). 



Example 487. 

The area of a rectangle is 168 cm 2 . The perimeter of the same rectangle is 52 cm. 
What are the dimensions of the rectangle? 



x 



y 

xy = 168 

2x + 2y = 52 
-2x -2x 

2y = -2x + 52 



We don't know anything about length or width 

Use two variables, x and y 

Length times width gives the area. 

Also use perimeter formula. 

Solve by substitution, isolate y 

Divide each term by 2 



359 



2 2 2 

y = — x + 26 Substitute into area equation 

x(-x + 26) = 168 Distribute 

— x 2 + 26x = 168 Divide each term by — 1 , changing all the signs 

x 2 — 2Qx = — 168 Solve by completing the square. 

1 V (\ x2 

— •26] =13 2 = 169 Find number to complete the square: f— ■ b 

x 2 - 2Qx + 324 = 1 Add 169 to both sides 

(x-13) 2 = l Factor 

x — 13 = ±1 Square root both sides 
+ 13 +13 

x = 13 ± 1 Evaluate 

x = 14 or 12 Two options for first side. 

y = - (14) + 26 = 12 Substitute 14 into y = - x + 26 

y = - (12) + 26 = 14 Substitute 12 into y = - x + 26 

Both are the same rectangle, variables switched! 

1 2 cm by 14 cm Our Solution 

World View Note: Indian mathematical records from the 9th century demon- 
strate that their civilization had worked extensivly in geometry creating religious 
alters of various shapes including rectangles. 

Another type of rectangle problem is what we will call a "frame problem". The 
idea behind a frame problem is that a rectangle, such as a photograph, is centered 
inside another rectangle, such as a frame. In these cases it will be important to 
rememember that the frame extends on all sides of the rectangle. This is shown in 
the following example. 

Example 488. 

An 8 in by 12 in picture has a frame of uniform width around it. The area of the 
frame is equal to the area of the picture. What is the width of the frame? 

Draw picture, picture if 8 by 10 

If frame has width x , on both sides, we add 2x 

2x 

8-12 = 96 Area of the picture, length times width 

2-96 = 192 Frame is the same as the picture. Totalareaisdoublethis. 

(12 + 2x) (8 + 2x) = 192 Area of everything, length times width 

96 + 24x + 16x + 4a; 2 =192 FOIL 

4x 2 + 40x + 96 = 192 Combine like terms 

— 192 — 192 Make equation equal to zero by subtracting 192 

4a; 2 + 40x - 96 = Factor out GCF of 4 



360 




A(x 2 + 10a; - 24) = 

4(a;-2)(a; + 12)=0 

a; - 2 = or £ + 12 = 

+ 2 + 2 -12-12 

x = 2 or -12 

2 inches 



Factor trinomial 

Set each factor equal to zero 

Solve each equation 

Can't have negative frame width. 
Our Solution 



Example 489. 

A farmer has a field that is 400 rods by 200 rods. He is mowing the field in a 
spiral pattern, starting from the outside and working in towards the center. After 
an hour of work, 72% of the field is left uncut. What is the size of the ring cut 
around the outside? 





400- 


-2x 




-2x 






200- 











200 



400 

400-200 = 80000 

80000 -(0.72) =57600 

(400 - 2a;) (200 - 2x) = 57600 

80000 - 800a; - 400x + 4x 2 = 57600 

4x 2 - 1200a; + 80000 = 57600 

- 57600 - 57600 

4x 2 - 1200a; + 22400 = 

4(x 2 - 300a; + 5600) =0 

4(a;-280)(x-20)=0 

a; -280 = or a; -20 = 

+ 280 + 280 +20 + 20 

x = 280 or 20 

20 rods 



Draw picture, outside is 200 by 400 
If frame has width x on both sides, 
subtract 2a; from each side to get center 



Area of entire field, length times width 

Area of center, multiply by 28% as decimal 

Area of center, length times width 

FOIL 

Combine like terms 

Make equation equal zero 

Factor out GCF of 4 

Factor trinomial 

Set each factor equal to zero 

Solve each equation 

The field is only 200 rods wide, 
Can't cut 280 off two sides! 
Our Solution 



For each of the frame problems above we could have also completed the square or 
use the quadratic formula to solve the trinomials. Remember that completing the 
square or the quadratic formula always will work when solving, however, factoring 
only works if we can factor the trinomial. 



361 



9.7 Practice - Rectangles 



1) In a landscape plan, a rectangular flowerbed is designed to be 4 meters longer 

than it is wide. If 60 square meters are needed for the plants in the bed, what 
should the dimensions of the rectangular bed be? 

2) If the side of a square is increased by 5 the area is multiplied by 4. Find the 

side of the original square. 

3) A rectangular lot is 20 yards longer than it is wide and its area is 2400 square 

yards. Find the dimensions of the lot. 

4) The length of a room is 8 ft greater than it is width. If each dimension is 
increased by 2 ft, the area will be increased by 60 sq. ft. Find the dimensions 
of the rooms. 

5) The length of a rectangular lot is 4 rods greater than its width, and its area is 

60 square rods. Find the dimensions of the lot. 

6) The length of a rectangle is 15 ft greater than its width. If each dimension is 

decreased by 2 ft, the area will be decreased by 106 ft 2 . Find the dimensions. 

7) A rectangular piece of paper is twice as long as a square piece and 3 inches 

wider. The area of the rectangular piece is 108 in 2 . Find the dimensions of the 
square piece. 

8) A room is one yard longer than it is wide. At 75c per sq. yd. a covering for 

the floor costs $31.50. Find the dimensions of the floor. 

9) The area of a rectangle is 48 ft 2 and its perimeter is 32 ft. Find its length and 

width. 

10) The dimensions of a picture inside a frame of uniform width are 12 by 16 
inches. If the whole area (picture and frame) is 288 in 2 , what is the width of 
the frame? 

11) A mirror 14 inches by 15 inches has a frame of uniform width. If the area of 
the frame equals that of the mirror, what is the width of the frame. 

12) A lawn is 60 ft by 80 ft. How wide a strip must be cut around it when 
mowing the grass to have cut half of it. 

13) A grass plot 9 yards long and 6 yards wide has a path of uniform width 
around it. If the area of the path is equal to the area of the plot, determine 
the width of the path. 

14) A landscape architect is designing a rectangular flowerbed to be border with 
28 plants that are placed 1 meter apart. He needs an inner rectangular space 
in the center for plants that must be 1 meter from the border of the bed and 

362 



that require 24 square meters for planting. What should the overall 
dimensions of the flowerbed be? 

15) A page is to have a margin of 1 inch, and is to contain 35 in 2 of painting. 
How large must the page be if the length is to exceed the width by 2 inches? 

16) A picture 10 inches long by 8 inches wide has a frame whose area is one half 
the area of the picture. What are the outside dimensions of the frame? 

17) A rectangular wheat field is 80 rods long by 60 rods wide. A strip of uniform 
width is cut around the field, so that half the grain is left standing in the form 
of a rectangular plot. How wide is the strip that is cut? 

18) A picture 8 inches by 12 inches is placed in a frame of uniform width. If the 
area of the frame equals the area of the picture find the width of the frame. 

19) A rectangular field 225 ft by 120 ft has a ring of uniform width cut around 
the outside edge. The ring leaves 65% of the field uncut in the center. What 
is the width of the ring? 

20) One Saturday morning George goes out to cut his lot that is 100 ft by 120 ft. 
He starts cutting around the outside boundary spiraling around towards the 
center. By noon he has cut 60% of the lawn. What is the width of the ring 
that he has cut? 

21) A frame is 15 in by 25 in and is of uniform width. The inside of the frame 
leaves 75% of the total area available for the picture. What is the width of the 
frame? 

22) A farmer has a field 180 ft by 240 ft. He wants to increase the area of the 
field by 50% by cultivating a band of uniform width around the outside. How 
wide a band should he cultivate? 

23) The farmer in the previous problem has a neighber who has a field 325 ft by 
420 ft. His neighbor wants to increase the size of his field by 20% by 
cultivating a band of uniform width around the outside of his lot. How wide a 
band should his neighbor cultivate? 

24) A third farmer has a field that is 500 ft by 550 ft. He wants to increase his 
field by 20%. How wide a ring should he cultivate around the outside of his 
field? 

25) Donna has a garden that is 30 ft by 36 ft. She wants to increase the size of 
the garden by 40%. How wide a ring around the outside should she cultivate? 

26) A picture is 12 in by 25 in and is surrounded by a frame of uniform width. 
The area of the frame is 30% of the area of the picture. How wide is the 
frame? 



363 



9.8 

Quadratics - Teamwork 



Objective: Solve teamwork problems by creating a rational equation to 
model the problem. 

If it takes one person 4 hours to paint a room and another person 12 hours to 
paint the same room, working together they could paint the room even quicker, it 
turns out they would paint the room in 3 hours together. This can be reasoned by 
the following logic, if the first person paints the room in 4 hours, she paints -j of 
the room each hour. If the second person takes 12 hours to paint the room, he 
paints — of the room each hour. So together, each hour they paint - + — of the 
room. Using a common denominator of 12 gives: — + — = — = -. This means 

-L — J_ ^ J. Z, O 

each hour, working together they complete - of the room. If - is completed each 
hour, it follows that it will take 3 hours to complete the entire room. 



This pattern is used to solve teamwork problems. If the first person does a job in 
A, a second person does a job in B, and together they can do a job in T (total). 
We can use the team work equation. 

Teamwork Equation: — - + — = — 
H A B T 

Often these problems will involve fractions. Rather than thinking of the first frac- 
tion as — , it may be better to think of it as the reciprocal of A's time. 

World View Note: When the Egyptians, who were the first to work with frac- 
tions, wrote fractions, they were all unit fractions (numerator of one). They only 
used these type of fractions for about 2000 years! Some believe that this cumber- 
some style of using fractions was used for so long out of tradition, others believe 
the Egyptians had a way of thinking about and working with fractions that has 
been completely lost in history. 



Example 490. 

Adam can clean a room in 3 hours. If his sister Maria helps, they can clean it in 
2- hours. How long will it take Maria to do the job alone? 

2 12 2 

2— = — Together time, 2—, needs to be converted to fraction 

5 5" 5 

5 
Adan: 3, Maria: x, Total: — Clearly state times for each and total, using x for Maria 



364 



115 

— -\ — = — Using reciprocals, add the individual times gives total 

3 x 12 

M1M + MlM = MlM Multiply each term by LCD of 12z 

4x + 12 = 5x Reduce each fraction 

— 4x — 4x Move variables to one side, subtracting 4x 

12 = a; Our solution for x 

It takes Maria 12 hours Our Solution 



Somtimes we only know how two people's times are related to eachother as in the 
next example. 

Example 491. 

Mike takes twice as long as Rachel to complete a project. Together they can com- 
plete the project in 10 hours. How long will it take each of them to complete the 
project alone? 

Mike: 2x, Rachel: x, Total: 10 Clearly define variables. If Rachelisx, Mike is 2x 

- — | — = — - Using reciprocals, add individal times equaling total 
2x x 10 

Ijl^ + 1CL0^ = 1(10^ Multiply each term by LCD, lOz 
2x x 10 

5 + 10 = 2 Combine like terms 

15 = x We have our x , we said x was Rachel's time 

2(15) = 30 Mike is double Rachel, this gives Mike's time. 

Mike: 30 hr, Rachel: 15 hr Our Solution 



With problems such as these we will often end up with a quadratic to solve. 



Example 492. 

Brittney can build a large shed in 10 days less than Cosmo can. If they built it 
together it would take them 12 days. How long would it take each of them 
working alone? 

Britney: x — 10, Cosmo: x, Total: 12 If Cosmo is x , Britney is x — 10 

— -\ — = — Using reciprocals, make equation 

x — 10 x 12 



365 



l(12x(x - 10)) l(12x(x - 10)) _ l(12x(x - 10)) 
x-10 ' x ~~ 12 



Multiply by LCD: 12x(x - 10) 



12x + 12(x — 10) =x(x — 10) Reduce fraction 

12x + 12x-120 = x 2 -10x Distribute 

24x — 120 = x 2 — lOx Combine like terms 

-24x + 120 -24x + 120 Move all terms to one side 

= x 2 -34x + 120 Factor 

= (x — 30) (x — 4) Set each factor equal to zero 

x — 30 = or x — 4 = Solve each equation 
+ 30 + 30 +4 + 4 



a; = 30 or x = 4 This,x, was defined as Cosmo. 
30 — 10 = 20 or 4 — 10 = — 6 Find Britney, can't have negative time 
Britney: 20 days, Cosmo: 30 days Our Solution 



In the previous example, when solving, one of the possible times ended up nega- 
tive. We can't have a negative amount of time to build a shed, so this possibility 
is ignored for this problem. Also, as we were solving, we had to factor x 2 — 34x + 
120. This may have been difficult to factor. We could have also chosen to com- 
plete the square or use the quadratic formula to find our solutions. 

It is important that units match as we solve problems. This means we may have 
to convert minutes into hours to match the other units given in the problem. 

Example 493. 

An electrician can complete a job in one hour less than his apprentice. Together 
they do the job in 1 hour and 12 minutes. How long would it take each of them 
working alone? 

12 
1 hr 12 min = 1 — hr Change 1 hour 12 minutes to mixed number 
60 

1— - = 1— = — Reduce and convert to fraction 
60 5 5 

6 
Electrician: x — 1, Apprentice: x, Total: — Clearly define variables 

H — = — Using reciprocals, make equation 



x — 1 x 6 



1(6 ^- 1)} + 1(6x( ^ 1)) = 5(6x( 6 X " l) Multiply each term by LCD 6x(x - 1) 



366 



6x + 6(x — 1) = 
62 + 6x — 6 
12a; -6 
-12a; + 6 



5a;(a; — 1) 
= 5a; 2 — 5a: 
= 5a; 2 — 5a; 
-12a; + 6 



5x 



2 

5 



= 5a; 2 -17a; + 6 
= (5a;-2)(x-3) 
-2 = or a: -3 = 

-2 + 2 +3 + 3 

= 2 or x = 3 

2 
"5 
or 3 



5x 



1 



x - 

-3 

5 



or a; : 



1 = 2 



Reduce each fraction 

Distribute 

Combine like terms 

Move all terms to one side of equation 

Factor 

Set each factor equal to zero 

Solve each equation 



Electrician: 2 hr, Apprentice: 3 hours 



Subtract 1 from each to find electrician 

Ignore negative. 
Our Solution 



Very similar to a teamwork problem is when the two involved parts are working 
against each other. A common example of this is a sink that is filled by a pipe 
and emptied by a drain. If they are working against eachother we need to make 
one of the values negative to show they oppose eachother. This is shown in the 
next example.. 

Example 494. 

A sink can be filled by a pipe in 5 minutes but it takes 7 minutes to drain a full 
sink. If both the pipe and the drain are open, how long will it take to fill the 
sink? 



Sink: 5, Drain: 7, Total: x Define variables, drain is negative 

1 1 1 

— — — = — U sing reciprocals to make equation, 

5 7 x 

Subtract because they are opposite 
l(35x) l(35.r) _ l(35ar) Multiplyeachtermby LCD: S5x 



7 



./• 



7 a; — 5a; = 35 
2a; = 35 

a;=17.5 
17.5 minor 17min 30 sec 



Reduce fractions 
Combine like terms 
Divide each term by 2 
Our answer for x 
Our Solution 



367 



9.8 Practice - Teamwork 

1) Bills father can paint a room in two hours less than Bill can paint it. Working 
together they can complete the job in two hours and 24 minutes. How much 
time would each require working alone? 

2) Of two inlet pipes, the smaller pipe takes four hours longer than the larger pipe 
to fill a pool. When both pipes are open, the pool is filled in three hours and 
forty-five minutes. If only the larger pipe is open, how many hours are required 
to fill the pool? 

3) Jack can wash and wax the family car in one hour less than Bob can. The two 
working together can complete the job in 1 - hours. How much time would 
each require if they worked alone? 

4) If A can do a piece of work alone in 6 days and B can do it alone in 4 days, 
how long will it take the two working together to complete the job? 

5) Working alone it takes John 8 hours longer than Carlos to do a job. Working 
together they can do the job in 3 hours. How long will it take each to do the 
job working alone? 

6) A can do a piece of work in 3 days, B in 4 days, and C in 5 days each working 
alone. How long will it take them to do it working together? 

7) A can do a piece of work in 4 days and B can do it in half the time. How long 
will it take them to do the work together? 

8) A cistern can be filled by one pipe in 20 minutes and by another in 30 minutes. 
How long will it take both pipes together to fill the tank? 

9) If A can do a piece of work in 24 days and A and B together can do it in 6 
days, how long would it take B to do the work alone? 

10) A carpenter and his assistant can do a piece of work in 3- days. If the 
carpenter himself could do the work alone in 5 days, how long would the 
assistant take to do the work alone? 

11) If Sam can do a certain job in 3 days, while it takes Fred 6 days to do the 
same job, how long will it take them, working together, to complete the job? 

12) Tim can finish a certain job in 10 hours. It take his wife JoAnn only 8 hours 
to do the same job. If they work together, how long will it take them to 
complete the job? 

13) Two people working together can complete a job in 6 hours. If one of them 
works twice as fast as the other, how long would it take the faster person, 
working alone, to do the job? 

14) If two people working together can do a job in 3 hours, how long will it take 
the slower person to do the same job if one of them is 3 times as fast as the 
other? 

15) A water tank can be filled by an inlet pipe in 8 hours. It takes twice that long 
for the outlet pipe to empty the tank. How long will it take to fill the tank if 
both pipes are open? 

368 



16) A sink can be filled from the faucet in 5 minutes. It takes only 3 minutes to 
empty the sink when the drain is open. If the sink is full and both the faucet 
and the drain are open, how long will it take to empty the sink? 

17) It takes 10 hours to fill a pool with the inlet pipe. It can be emptied in 15 hrs 
with the outlet pipe. If the pool is half full to begin with, how long will it 
take to fill it from there if both pipes are open? 

18) A sink is - full when both the faucet and the drain are opened. The faucet 
alone can fill the sink in 6 minutes, while it takes 8 minutes to empty it with 

Q 

the drain. How long will it take to fill the remaining - of the sink? 

19) A sink has two faucets, one for hot water and one for cold water. The sink 
can be filled by a cold-water faucet in 3.5 minutes. If both faucets are open, 
the sink is filled in 2.1 minutes. How long does it take to fill the sink with just 
the hot-water faucet open? 

20) A water tank is being filled by two inlet pipes. Pipe A can fill the tank in 4- 
hrs, while both pipes together can fill the tank in 2 hours. How long does it 
take to fill the tank using only pipe B? 

21) A tank can be emptied by any one of three caps. The first can empty the 
tank in 20 minutes while the second takes 32 minutes. If all three working 
together could empty the tank in 8— minutes, how long would the third take 
to empty the tank? 

22) One pipe can fill a cistern in 1- hours while a second pipe can fill it in 2- hrs. 
Three pipes working together fill the cistern in 42 minutes. How long would it 
take the third pipe alone to fill the tank? 

23) Sam takes 6 hours longer than Susan to wax a floor. Working together they 
can wax the floor in 4 hours. How long will it take each of them working 
alone to wax the floor? 

24) It takes Robert 9 hours longer than Paul to rapair a transmission. If it takes 
them 2- hours to do the job if they work together, how long will it take each 
of them working alone? 

25) It takes Sally 10- minutes longer than Patricia to clean up their dorm room. 
If they work together they can clean it in 5 minutes. How long will it take 
each of them if they work alone? 

26) A takes 7- minutes longer than B to do a job. Working together they can do 
the job in 9 minutes. How long does it take each working alone? 

27) Secretary A takes 6 minutes longer than Secretary B to type 10 pages of 

Q 

manuscript. If they divide the job and work together it will take them 8- 
minutes to type 10 pages. How long will it take each working alone to type 
the 10 pages? 

28) It takes John 24 minutes longer than Sally to mow the lawn. If they work 
together they can mow the lawn in 9 minutes. How long will it take each to 
mow the lawn if they work alone? 



369 



9.9 

Quadratics - Simultaneous Products 



Objective: Solve simultaneous product equations using substitution to 
create a rational equation. 

When solving a system of equations where the variables are multiplied together 
we can use the same idea of substitution that we used with linear equations. 
When we do so we may end up with a quadratic equation to solve. When we used 
substitution we solved for a variable and substitute this expression into the other 
equation. If we have two products we will choose a variable to solve for first and 
divide both sides of the equations by that variable or the factor containing the 
variable. This will create a situation where substitution can easily be done. 

Example 495. 

x y = 48 To solve for x , divide fir st 

(x + 3)(y — 2) = 54 equation by x , second by x + 3 

y = — and y — 2 = Substitute — for y in the second equation 

X X ~y~ O X 

48 54 

2 = Multiply each term by LCD: x(x + 3) 

x x + 3 

48x(x + 3) _. , oN 54x(x + 3) _ , . r 

— 2x(x + 3) = Reduce each traction 

x v ; x+3 

48(x + 3) -2x(x + 3) = 54x Distribute 

48x + 144 — 2x 2 — 6x = 54x Combine like terms 

— 2x 2 + 42x + 144 = 54x Make equation equal zero 

— 54x — 54x Subtract 54x from both sides 

- 2x 2 - 12x + 144 = Divide each term by GCF of - 2 

x 2 + 6x - 72 = Factor 

(x — 6) (x + 12) = Set each factor equal to zero 

x — 6 = or x + 12 = Solve each equation 
+ 6 + 6 -12-12 

x = 6 or x = — 12 Substitute each solution into xy = 48 

6y = 48 or — 12y = 48 Solve each equation 



6 6 -12-12 

y = 8 or y = — 4 Our solutions for y , 
(6,8) or ( — 12,-4) Our Solutions as ordered pairs 



370 



These simultaneous product equations will also solve by the exact same pattern. 
We pick a variable to solve for, divide each side by that variable, or factor con- 
taining the variable. This will allow us to use substitution to create a rational 
expression we can use to solve. Quite often these problems will have two solu- 
tions. 



Exampl 


e 496 












xy = — 35 








(a + 6)0 


y-2)-. 


= 5 




— 


35 

rinrl rt i 


o — 


5 




y — 


x y 


z — 
X 


+ 6 






-35 

X 


9 — 


5 




X 


+ 6 


X 


+ 6) 


■ 2x(x + 6) = 


5x(x + 6) 
x + 6 




-35(x + 6)-2x(a 


; + 6) = 


= 5:r 




— 35a; 


- 210 -2a; 2 - 


-12x = 


= 5a: 






- 2x 2 - 47 'x - 


-210 = 


= 5x 






— 5x 




- 5x 






- 2x 2 - 52a 


-210 


= 






x 2 + 26a: + 105 


= 






(x + 5)(x + 21) 


= 




x + 5 = or 


x + 21 


= 






-5-5 


-21- 


-21 






x = — 5 oi 


: x = - 


-21 




-5y = 


— 35 or — 21y = - 


-35 




^5 


-5 


21 - 


-21 

5 






y = l 


or y 


~3 




( 


-5,7) or 1 


-21, 


!) 



To solve for x , divide the first 
equation by x , second by x + 6 

— 35 

Substitute for y in the second equation 

Multiply each term by LCD: x(x + 6) 

Reduce fractions 

Distribute 
Combine like terms 
Make equation equal zero 

Divide each term by — 2 

Factor 

Set each factor equal to zero 

Solve each equation 

Substitute each solution into x y = — 35 
Solve each equation 

Our solutions for y 

Our Solutions as ordered pairs 



The processes used here will be used as we solve applications of quadratics 
including distance problems and revenue problems. These will be covered in 
another section. 

World View Note: William Horner, a British mathematician from the late 18th 
century/early 19th century is credited with a method for solving simultaneous 
equations, however, Chinese mathematician Chu Shih-chieh in 1303 solved these 
equations with exponents as high as 14! 



371 



9.9 Practice - Simultaneous Product 



Solve. 

I) xy = 72 

(x + 2)(y-4) = 128 

3) xy = 150 

(x-6)(y + l)=64 

5) xy = 45 

(x + 2)(y + l)=70 

7) xy = 90 

(x-5)(y + l) = 120 

9) xy = 12 

(x+l)(y-4) = 16 

II) xy = 45 
(x-5)(y + 3) = 160 



2) xy = 180 

(x-l)(y-±) = 205 

4) xy = 120 

(x + 2)(y-3) = 120 

6) xy = 65 

(x-8)(y + 2) = 35 

8) xy = 48 

(x-6)(y + 3) = 60 

10) xy = 60 

(x + 5)(y + 3) = 150 

12) xy = 80 

(x-5)(y + 5) = 45 



372 



9.10 



Quadratics - Revenue and Distance 



Objective: Solve revenue and distance applications of quadratic equa- 
tions. 

A common application of quadratics comes from revenue and distance problems. 
Both are set up almost identical to each other so they are both included together. 
Once they are set up, we will solve them in exactly the same way we solved the 
simultaneous product equations. 

Revenue problems are problems where a person buys a certain number of items 
for a certain price per item. If we multiply the number of items by the price per 
item we will get the total paid. To help us organize our information we will use 
the following table for revenue problems 





Number 


Price 


Total 


First 








Second 









The price column will be used for the individual prices, the total column is used 
for the total paid, which is calculated by multiplying the number by the price. 
Once we have the table filled out we will have our equations which we can solve. 
This is shown in the following examples. 

Example 497. 

A man buys several fish for $56. After three fish die, he decides to sell the rest at 
a profit of $5 per fish. His total profit was $4. How many fish did he buy to 
begin with? 





Number 


Price 


Total 


Buy 


n 


P 


56 


Sell 









Using our table, we don't know the number 
he bought, or at what price, so we use varibles 
n and p. Total price was $56. 





Number 


Price 


Total 


Buy 


n 


P 


56 


Sell 


n — 3 


p + 5 


60 



When he sold, he sold 3 less (n — 3), for $5 
more (p + 5) . Total profit was $4, combined 
with $56 spent is $60 



np = 56 Find equatinos by multiplying number by price 
(n — 3) (p + 5) = 60 These are a simultaneous product 



56 60 

p = — and p + 5 = - 

n n — 6 



Solving for number, divide by nor (n — 3) 



373 



56 r 60 

— + 5 = 

n n — 3 



56 
Substitute — for p in second equation 
n 



- + 5n(n — 3) = — - — - Multiply each term by LCD: n(n — 3) 



n 



n — 3 



n ■ 



56(n — 3) + 5n(n — 3) = 60n 

56n — 168 + 5n 2 — 15n = 60n 

5n 2 + 41n-168 = 60n 

— 60n — 60n 

5n 2 -19n- 168 = 
19 ± ^(-IQ) 2 - 4(5) (-168) 
2(5) 

19±\/3721 19 ±61 



n 



10 



10 



n - 



80 
10 



Reduce fractions 
Combine like terms 
Move all terms to one side 

Solve with quadratic formula 
Simplify 

We don't want negative solutions, only do 



: 8 This is our n 



8 fish Our Solution 



Example 498. 

A group of students together bought a couch for their dorm that cost $96. How- 
ever, 2 students failed to pay their share, so each student had to pay $4 more. 
How many students were in the original group? 



$96 was paid, but we don't know the number 
or the price agreed upon by each student . 





Number 


Price 


Total 


Deal 


n 


P 


96 


Paid 











Number 


Price 


Total 


Deal 


n 


P 


96 


Paid 


n-2 


p + 4 


96 



There were 2 less that actually paid (n — 2) 
and they had to pay $4 more (p + 4) . The 
total here is still $96. 



np = 96 Equations are product of number and price 

(n — 2) (p + 4) = 96 This is a simultaneous product 

96 96 

p = — and p + 4 = — Solving for number, divide by nandn — 2 

n n — 2 

96 , 96 _ . . 96, . „, 

h 4 = — Substitute — tor p in the second equation 

n n — 2 n 



374 



96n(n-2) , . _ 96n(n - 2) 



n 







n - 


-2 


96(n 


-2)+4n(n 


-2) = 


= 96n 


96n 


-192 + 4n 2 - 


-8n = 


= 96n 




4n 2 + 88n - 


192 = 


= 96n 




— 96n 




- 96n 




4n 2 — 8n 


-192 = 




+ 192 + 192 



4n 2 -8n = 192 
n 2 -2n = 48 

2 



2- 



n 



2 
2 2n 



1 2 =1 



(n 



f 1=49 

1) 2 = 49 

n-l=±7 

n = l±7 

n=l+7=8 

8 students 



Multiply each term by LCD: n(n — 2) 

Reduce fractions 

Distribute 

Combine like terms 

Set equation equal to zero 

Solve by completing the square, 

Separate variables and constant 

Divide each term by a or 4 



Complete the square: I b- 



1 



Add to both sides of equation 

Factor 

Square root of both sides 

Add 1 to both sides 

We don't want a negative solution 

Our Solution 



The above examples were solved by the quadratic formula and completing the 
square. For either of these we could have used either method or even factoring. 
Remember we have several options for solving quadratics. Use the one that seems 
easiest for the problem. 

Distance problems work with the same ideas that the revenue problems work. 
The only difference is the variables are r and t (for rate and time), instead of 
n and p (for number and price). We already know that distance is calculated by 
multiplying rate by time. So for our distance problems our table becomes the fol- 
lowing: 





rate 


time 


distance 


First 








Second 









Using this table and the exact same patterns as the revenue problems is shown in 
the following example. 

Example 499. 



375 



Greg went to a conference in a city 120 miles away. On the way back, due to road 
construction he had to drive 10 mph slower which resulted in the return trip 
taking 2 hours longer. How fast did he drive on the way to the conference? 



We do not know rate, r , or time, t he traveled 
on the way to the conference. But we do know 
the distance was 120 miles. 





rate 


time 


distance 


There 


r 


t 


120 


Back 











rate 


time 


distance 


There 


r 


t 


120 


Back 


r-10 


t + 2 


120 



Coming back he drove 10 mph slower (r — 10) 
and took 2 hours longer (t + 2). The distance 
was still 120 miles. 



rt= 120 

(r- 10)(t + 2) = 120 



Equations are product of rate and time 
We have simultaneous product equations 



,120 . , „ 120 
t= and t + 2 



120r(r-10) 



120 



+ 2r(r-10) 



f2 



r-10 
120 



r-10 

120r(r-10) 

r-10 



Solving for rate, divide by r and r — 10 



120 

Substitute for t in the second equation 

r 



Multiply each term by LCD: r(r — 10) 



120(r - 10) + 2r 2 - 20r = 120r 

120r - 1200 + 2r 2 - 20r = 120r 

2r 2 + 100r-1200 = 120r 

-120r -120r 

2r 2_20r- 1200 = 

r 2 -10r- 600 = 

(r-30)(r + 20)=0 

r-30 = and r + 20 = 



+ 30 + 30 



20-20 



r = 30 and r = — 20 
30 mph 



Reduce each fraction 

Distribute 

Combine like terms 

Make equation equal to zero 

Divide each term by 2 

Factor 

Set each factor equal to zero 

Solve each equation 

Can't have a negative rate 
Our Solution 



World View Note: The world's fastest man (at the time of printing) is 
Jamaican Usain Bolt who set the record of running 100 m in 9.58 seconds on 
August 16, 2009 in Berlin. That is a speed of over 23 miles per hour! 

Another type of simultaneous product distance problem is where a boat is trav- 
eling in a river with the current or against the current (or an airplane flying with 
the wind or against the wind). If a boat is traveling downstream, the current will 
push it or increase the rate by the speed of the current. If a boat is traveling 



376 



upstream, the current will pull against it or decrease the rate by the speed of the 
current. This is demonstrated in the following example. 

Example 500. 

A man rows down stream for 30 miles then turns around and returns to his orig- 
inal location, the total trip took 8 hours. If the current flows at 2 miles per hour, 
how fast would the man row in still water? 





rate 


time 


distance 


down 




t 


30 


up 




8-t 


30 



Write total time above time column 
We know the distance up and down is 30. 
Put t for time downstream. Subtracting 
8 — t becomes time upstream 





rate 


time 


distance 


down 


r + 2 


t 


30 


up 


r-2 


8-t 


30 



Downstream the current of 2 mph pushes 
the boat (r + 2) and upstream the current 
pulls the boat (r — 2) 



(r + 2)i = 30 Multiply rate by time to get equations 
(r — 2) (8 — t) = 30 We have a simultaneous product 



30 30 

t = and 8 — t = Solving for rate, divide by r + 2orr — 2 



r + 2 



r-2 



8 = Substitute for t in second equation 

r+2 r-2 r+2 H 



(r + 2)(r-2)- 



30(r + 2) (r - 2) _ 30(r + 2) (r - 2) 



r + 2 



r-2 



Multiply each term by LCD: (r + 2) (r - 2) 



](r + 2)(r - 2) - 30(r - 2) = 30(r + 2) 

8r 2 - 32 - 30r + 60 = 30r + 60 

8r 2 -30r + 28 = 30r + 60 

- 30r - 60 - 30r - 60 



Reduce fractions 
Multiply and distribute 
Make equation equal zero 



8r 2 - 60r - 32 = 


Divide each term by 4 


2r 2 -15r-8 = 


Factor 


(2r + l)(r-8) = 


Set each factor equal to zero 


r + l = or r-8 = 


Solve each equation 


-1-1 +8+8 




2r = — 1 or r = 8 




~T ~2~ 

1 





or r : 



■ mph 



Can't have a negative rate 
Our Solution 



377 



9.10 Practice - Revenue and Distance 



1) A merchant bought some pieces of silk for $900. Had he bought 3 pieces more 
for the same money, he would have paid $15 less for each piece. Find the 
number of pieces purchased. 

2) A number of men subscribed a certain amount to make up a deficit of $100 
but 5 men failed to pay and thus increased the share of the others by $1 each. 
Find the amount that each man paid. 

3) A merchant bought a number of barrels of apples for $120. He kept two barrels 
and sold the remainder at a profit of $2 per barrel making a total profit of $34. 
How many barrels did he originally buy? 

4) A dealer bought a number of sheep for $440. After 5 had died he sold the 
remainder at a profit of $2 each making a profit of $60 for the sheep. How 
many sheep did he originally purchase? 

5) A man bought a number of articles at equal cost for $500. He sold all but two 
for $540 at a profit of $5 for each item. How many articles did he buy? 

6) A clothier bought a lot of suits for $750. He sold all but 3 of them for $864 
making a profit of $7 on each suit sold. How many suits did he buy? 

7) A group of boys bought a boat for $450. Five boys failed to pay their share, 
hence each remaining boys were compelled to pay $4.50 more. How many boys 
were in the original group and how much had each agreed to pay? 

8) The total expenses of a camping party were $72. If there had been 3 fewer 
persons in the party, it would have cost each person $2 more than it did. How 
many people were in the party and how much did it cost each one? 

9) A factory tests the road performance of new model cars by driving them at two 
different rates of speed for at least 100 kilometers at each rate. The speed rates 
range from 50 to 70 km/hr in the lower range and from 70 to 90 km/hr in the 
higher range. A driver plans to test a car on an available speedway by driving 
it for 120 kilometers at a speed in the lower range and then driving 120 
kilometers at a rate that is 20 km/hr faster. At what rates should he drive if he 
plans to complete the test in 3- hours? 

10) A train traveled 240 kilometers at a certain speed. When the engine was 
replaced by an improved model, the speed was increased by 20 km/hr and the 
travel time for the trip was decreased by 1 hour. What was the rate of each 
engine? 

11) The rate of the current in a stream is 3 km/hr. A man rowed upstream for 3 
kilometers and then returned. The round trip required 1 hour and 20 minutes. 
How fast was he rowing? 

378 



12) A pilot flying at a constant rate against a headwind of 50 km/hr flew for 750 
kilometers, then reversed direction and returned to his starting point. He 
completed the round trip in 8 hours. What was the speed of the plane? 

13) Two drivers are testing the same model car at speeds that differ by 20 km/hr. 
The one driving at the slower rate drives 70 kilometers down a speedway and 

returns by the same route. The one driving at the faster rate drives 76 
kilometers down the speedway and returns by the same route. Both drivers 
leave at the same time, and the faster car returns - hour earlier than the 
slower car. At what rates were the cars driven? 

14) An athlete plans to row upstream a distance of 2 kilometers and then return 
to his starting point in a total time of 2 hours and 20 minutes. If the rate of 
the current is 2 km/hr, how fast should he row? 

15) An automobile goes to a place 72 miles away and then returns, the round 
trip occupying 9 hours. His speed in returning is 12 miles per hour faster than 
his speed in going. Find the rate of speed in both going and returning. 

16) An automobile made a trip of 120 miles and then returned, the round trip 
occupying 7 hours. Returning, the rate was increased 10 miles an hour. Find 
the rate of each. 

17) The rate of a stream is 3 miles an hour. If a crew rows downstream for a 
distance of 8 miles and then back again, the round trip occupying 5 hours, 
what is the rate of the crew in still water? 

18) The railroad distance between two towns is 240 miles. If the speed of a train 
were increased 4 miles an hour, the trip would take 40 minutes less. What is 
the usual rate of the train? 

19) By going 15 miles per hour faster, a train would have required 1 hour less to 
travel 180 miles. How fast did it travel? 

20) Mr. Jones visits his grandmother who lives 100 miles away on a regular basis. 
Recently a new freeway has opend up and, although the freeway route is 120 
miles, he can drive 20 mph faster on average and takes 30 minutes less time to 
make the trip. What is Mr. Jones rate on both the old route and on the 

freeway? 

21) If a train had traveled 5 miles an hour faster, it would have needed 1 - hours 
less time to travel 150 miles. Find the rate of the train. 

22) A traveler having 18 miles to go, calculates that his usual rate would make 
him one-half hour late for an appointment; he finds that in order to arrive on 
time he must travel at a rate one-half mile an hour faster. What is his usual 
rate? 



379 



9.11 



Quadratics - Graphs of Quadratics 



Objective: Graph quadratic equations using the vertex, x-intercepts, 
and y-intercept. 

Just as we drew pictures of the solutions for lines or linear equations, we can draw 
a picture of solution to quadratics as well. One way we can do that is to make a 
table of values. 



Example 501. 



y = x 2 — 4x + 3 Make a table of values 



X 


y 







1 




2 




3 




4 





y = (0) 2 + 4(0) + 3 = 0-0 + 3 = 3 

y = (l) 2 - 4(1) + 3 = 1 - 4 + 3 = 

y = (2) 2 - 4(2) + 3 = 4-8 + 3 = -l 

y = (3) 2 - 4(3) + 3 = 9- 12 + 3 = 

(4) 2 - 4(4) + 3 = 16 -16 + 3 = 3 



IJ 



We will test 5 values to get an idea of shape 



Plug in for x and evaluate 
Plug 1 in for x and evaluate 
Plug 2 in for x and evaluate 
Plug 3 in for x and evaluate 
Plug 4 in for x and evaluate 



Our completed table. Plot points on graph 



Plot the points (0, 3), (1, 0), (2, - 1), 
(3,0), and (4, 3). 

Connect the dots with a smooth 
curve. 

Our Solution 



When we have x 2 in our equations, the graph will no longer be a straight line. 
Quadratics have a graph that looks like a U shape that is called a parabola. 

World View Note: The first major female mathematician was Hypatia of Egypt 
who was born around 370 AD. She studied conic sections. The parabola is one 
type of conic section. 



X 


y 





3 


1 





2 


-1 


3 





4 


3 



m 



380 



The above method to graph a parabola works for any equation, however, it can be 
very tedious to find all the correct points to get the correct bend and shape. For 
this reason we identify several key points on a graph and in the equation to help 
us graph parabolas more efficiently. These key points are described below. 

Point A: y-intercept: Where the graph 
crosses the vertical y-axis. 

Points B and C: x-intercepts: Where 
the graph crosses the horizontal x-axis 

Point D: Vertex: The point where the 
graph curves and changes directions. 

We will use the following method to find each of the points on our parabola. 
To graph the equation y = a x 2 + b x + c, find the following points 

1. y-intercept: Found by making 2 = 0, this simplifies down to y = c 

2. x-intercepts: Found by making y = 0, this means solving = ax 2 + bx + c 

b 




3. Vertex: Let x ■ 



2,1 



to find x. Then plug this value into the equation to find 



After finding these points we can connect the dots with a smooth curve to find 
our graph! 

Example 502. 

y = x 2 + 4x + 3 Find the key points 

y = 3 y = c is the y — intercept 



= x 2 + 4x + 3 To find x — intercept we solve the equation 

= (x + 3)(x + l) Factor 

x + 3 = and x + 1 = Set each factor equal to zero 

— 3 — 3 —1 — 1 Solve each equation 



x 



3 and x = — 1 Our x — intercepts 



4 



x ■■ 



-4 -b 

— = — 2 To find the vertex, first use a; = 

2 ' 2a 



2(1) 

y = ( — 2) 2 + 4( — 2) + 3 Plug this answer into equation to find y — coordinate 

y = 4 - 8 + 3 Evaluate 

y = — 1 The y — coordinate 

( — 2, — 1) Vertex as a point 



381 



7 



/ 



Graph the y-intercept at 3, the x- 
intercepts at — 3 and — 1, and the 
vertex at ( — 2, — 1). Connect the dots 
with a smooth curve in a U shape to 
get our parabola. 

Our Solution 



If the a in y = ax 2 + bx + c is a negative value, the parabola will end up being an 
upside-down U. The process to graph it is identical, we just need to be very 
careful of how our signs operate. Remember, if a is negative, then ax 2 will also be 
negative because we only square the x, not the a. 

Example 503. 

y = — 3x 2 + Ylx — 9 Find key points 



y 



9 y — intercept is y = c 



= - 3x 2 + 12a: - 9 

= -3(x 2 -4x + 3) 

= -3(a;-3)(x-l) 

x — 3 = and x — 1 = 

+3+3 +1+1 



To find x — intercept solve this equation 
Factor out GCF first, then factor rest 
Set each factor with a varaible equal to zero 
Solve each equation 



3 and x = 1 Our x — intercepts 



12 



12 



x - 



2(-3) -6 

-3(2) 2 + 12(2)- 

y = -3(4) + 24- 
y = -12 + 24 - 

V 



2 To find the vertex, first use x ■ 



9 
9 
9 

3 



(2,3) 



A 



2 a 



Plug this value into equation to find y 
Evaluate 

y — value of vertex 
Vertex as a point 



coordinate 



Graph the y-intercept at — 9, the x- 
intercepts at 3 and 1, and the vertex at 
(2, 3). Connect the dots with smooth 
curve in an upside-down U shape to 
get our parabola. 

Our Solution 



382 



It is important to remember the graph of all quadratics is a parabola with the 
same U shape (they could be upside-down). If you plot your points and we cannot 
connect them in the correct U shape then one of your points must be wrong. Go 
back and check your work to be sure they are correct! 

Just as all quadratics (equation with y = x 2 ) all have the same U-shape to them 
and all linear equations (equations such as y = x) have the same line shape when 
graphed, different equations have different shapes to them. Below are some 
common equations (some we have yet to cover!) with their graph shape drawn. 



Absolute Value 



Cubic 



\x\ 



y = x 




Quadratic 



Exponential 



y = x 



V = a 



< 



Square Root 
y = Vx 



Logarithmic 
y = log a £ 





383 



9.11 Practice - Graphs of Quadratics 

Find the vertex and intercepts of the following quadratics. Use this 
information to graph the quadratic. 

I) y = x 2 -2x-8 2) y = x 2 -2x-3 

3) y = 2x 2 - Ylx + 10 4) y = 2x 2 - \2x + 16 

5) y = - 2x 2 + 12x - 18 6) y = - 2x 2 + 12x - 10 

7) y = - 3x 2 + 24x - 45 8) y = - 3x 2 + 12x - 9 

9) y=-x 2 + Ax + h 10) y=-x 2 + Ax-3 

II) y = - x 2 + 6x - 5 12) y = - 2x 2 + 16x - 30 
13) y = - 2x 2 + 16x - 24 14) y = 2x 2 + Ax - 6 

15) y = 3s 2 + 12a; + 9 16) j/ = 5x 2 + 30s + 45 

17) y = 5x 2 - 40x + 75 18) y = 5x 2 + 20x + 15 

19) y = - 5x 2 - 60x - 175 20) y=- 5x 2 + 20x - 15 



384 



Chapter 10 : Functions 



10.1 Function Notation 386 

10.2 Operations on Functions 393 

10.3 Inverse Functions 401 

10.4 Exponential Functions 406 

10.5 Logarithmic Functions 410 

10.6 Application: Compound Interest 414 

10.7 Trigonometric Functions 420 

10.8 Inverse Trigonometric Functions 428 



385 



10.1 

Functions - Function Notation 



Objective: Identity functions and use correct notation to evaluate func- 
tions at numerical and variable values. 

There are many different types of equations that we can work with in algebra. An 
equation gives the relationship between variables and numbers. Examples of sev- 
eral relationships are below: 



(x-S) 2 (y + 2) 2 _, . 2 . „ . , n 

- — — - — - — — = 1 and y = x — 2x + 7 and \Jy + x — 7 = xy 
9 4 



There is a speical classification of relationships known as functions. Functions 
have at most one output for any input. Generally x is the variable that we plug 
into an equation and evaluate to find y. For this reason x is considered an input 
variable and y is considered an output variable. This means the definition of a 
function, in terms of equations in x and y could be said, for any x value there is at 
most one y value that corresponds with it. 

A great way to visualize this definition is to look at the graphs of a few relation- 
ships. Because x values are vertical lines we will draw a vertical line through the 
graph. If the vertical line crosses the graph more than once, that means we have 
too many possible y values. If the graph crosses the graph only once, then we say 
the relationship is a function. 



386 



Example 504. 

Which of the following graphs are graphs of functions? 




Drawing a vertical line 
through this graph will 
only cross the graph 
once, it is a function. 




Drawing a vertical line 
through this graph will 
cross the graph twice, 
once at top and once at 
bottom. This is not a 
function. 




Drawing a vertical line 
through this graph will 
cross the graph only 
once, it is a function. 



We can look at the above idea in an algebraic method by taking a relationship 
and solving it for y. If we have only one solution then it is a function. 



Example 505. 




Is 3x 2 — y = 5 a function? 


Solve the relation for y 


- 3x 2 - 3x 2 


Subtract 3x 2 from both sides 


- y = - Sx 2 + 5 


Divide each term by — 1 


^T ^T ^T 




y = 3x 2 - 5 


Only one solution for y. 



Yes! It is a function 



Example 506. 








Is y 2 — x = 5 a funct 


ion? 


Solve the relation for y 




+ X + X 




Add x to both sides 




y 2 = x + 5 




Square root of both sdies 
Simplify 




\Jy 2 = ± y/x + 5 




y = ± \J x + 5 




Two solutions for y (one + , one - 


") 




No! 


Not a function 





Once we know we have a function, often we will change the notation used to 
emphasis the fact that it is a function. Instead of writing y = , we will use func- 
tion notation which can be written f(x) = . We read this notation "/ of x". So for 



387 



the above example that was a function, instead of writing y = 3x 2 — 5, we could 
have written /(#) = 3x 2 — 5. It is important to point out that f(x) does not mean 
/ times x, it is mearly a notation that names the function with the first letter 
(function /) and then in parenthesis we are given information about what vari- 
ables are in the function (variable x). The first letter can be anything we want it 
to be, often you will see g(x) (read g of x). 

World View Note: The concept of a function was first introduced by Arab 
mathematician Sharaf al-Din al-Tusi in the late 12th century 

Once we know a relationship is a function, we may be interested in what values 
can be put into the equations. The values that are put into an equation (generally 
the x values) are called the domain. When finding the domain, often it is easier 
to consider what cannot happen in a given function, then exclude those values. 



Example 507. 



Find the domain: f(x) = 2 — — With fractions, zero can't be in denominator 



3x — 1 
x 2 + x — 6 

x 2 + x — 6 ^ Solve by factoring 

(x + 3) (x — 2) =^ Set each factor not equal to zero 

x + 3 ^ and x — 2 ^ Solve each equation 

-3-3 +2+2 



x^ — 3,2 Our Solution 



The notation in the previous example tells us that x can be any value except for 
— 3 and 2. If x were one of those two values, the function would be undefined. 



Example 508. 

Find the domain: f(x) = 3x 2 — x With this equation there are no bad values 
All Real Numbers or R Our Solution 



In the above example there are no real numbers that make the function unde- 
fined. This means any number can be used for x. 



Example 509. 

Find the domain: f(x) = \J1x — 3 Square roots can't be negative 

388 



2x - 3 ^ 

+ 3 + 3 

2x^3 

~2~ ~2~ 
.3 



Set up an inequality 
Solve 



Our Solution 



The notation in the above example states that our variable can be ^ or any 
number larger than -. But any number smaller would make the function unde- 
fined (without using imaginary numbers). 

Another use of function notation is to easily plug values into functions. If we want 
to substitute a variable for a value (or an expression) we simply replace the vari- 
able with what we want to plug in. This is shown in the following examples. 



Example 510. 



/(x) = 3x 2 -4x;find/(-2) 


Substitute — 2 in for x in the function 


/( _ 2 ) = 3(-2) 2 -4(-2) 


Evaluate, exponents first 


/(-2) = 3(4)-4(-2) 


Multiply 


/(-2) = 12 + 8 


Add 


/(-2) = 20 


Our Solution 



Example 511. 



h(x) 



:3 2x - 6 ;find/i(4) 

/ l (4)=3 2 W- 6 

/i(4) = 3 8 " 6 

/i(4) = 3 2 

/i(4) = 9 



Substitute 4 in for x in the function 
Simplify exponent, mutiplying first 
Subtract in exponent 
Evaluate exponent 
Our Solution 



Example 512. 



fc(o) = 2|o + 4|;findfc(-7) 

fc(-7) = 2|-7 + 4| 

Jfc(-7) = 2|-3| 

fc(-7) = 2(3) 



Substitute — 7 in for a in the function 
Add inside absolute values 
Evaluate absolute value 
Multiply 



389 



k( — 7) = 6 Our Solution 



As the above examples show, the function can take many different forms, but the 
pattern to evaluate the function is always the same, replace the variable with 
what is in parenthesis and simplify. We can also substitute expressions into func- 
tions using the same process. Often the expressions use the same variable, it is 
important to remember each variable is replaced by whatever is in parenthesis. 



Example 513. 



g (x) = x A + 1; find g(Sx) 

g(Sx) = (Sx) 4 + l 

g(3x)=81x 4 + l 



Replace x in the function with (3x) 
Simplify exponet 
Our Solution 



Example 514. 



p(t) = t 2 -t;tindp(t + l) 

p(t+l) = (t + l) 2 -(t + l) 

p{t + l)=t 2 + 2t+l-{t + l) 

p(t + I) = t 2 + 2t + 1 - t - I 

p{t+l) = t 2 + t 



Replace each t in p(t) with (t+1) 
Square binomial 
Distribute negative 
Combine like terms 
Our Solution 



It is important to become comfortable with function notation and how to use it as 
we transition into more advanced algebra topics. 



390 



10.1 Practice - Function Notation 



Solve. 

1) Which of the following is a function? 

a) b) 




c) 




e) y = 3x — 7 

g) Jy+x = 2 




d) 




f) y2 _ x 2 = 1 

h) x 2 + y 2 = 1 



Specify the domain of each of the following funcitons. 

2) f(x) = - 5x + 1 3) /(£) = \/5^4x 

4)«(t) = ^ 5) /(x)=s 2 -3x-4 

6)s(*) = ^r 7) f(x) = y/¥=W 



8) /(^) = a2 _ 3:c _ 4 



9) h(x) 



V3x-12 
x 2 -25 



391 



Evaluate each function. 



11 
13 

15 
17 
19 
21 
23 
25 
27 
29 
31 
33 
35 
37 
39 



g(x) = Ax - 4; Find g(0) 

/(x) = |3x+l| + l;Find/(0) 

f(n) = - 2| - n - 2| + 1; Find /( - 6) 

/(t) = 3*-2;Find/(-2) 

/(*) = l* + 3|;Find/(10) 

w(n) = 4n + 3; Find w(2) 

w(n)=2 n+2 ;Findw(-2) 

p(n) = — 3|n|; Findp(7) 

p(i) = -i 3 + i; Find p(4) 

fc(n) = |n — 1|; Find &(3) 

/i(a;) = x 3 + 2; Find h( - Ax) 

h(x) = Sx + 2; Find /i( - 1 + x) 

h(t) = 2| - 3* - l| + 2; Find h(n 2 ) 

g(x) = x + 1; Find g(3x) 

g(x) = 5 X ; Find g( — 3 — x) 



12 
14 
16 
18 
20 
22 
24 
26 
28 
30 
32 
34 
36 
38 
40 



g(n) 

f(n) 
f(a) 
w(x) 
w(x) 
p(x) 
k(a) 
k(x) 

P (t): 

h(n) 

h(a) 

h(x) 

h(t) 

h(n) 



:-3-5- n ;Find#(2) 
= x 2 + 4;Find/(-9) 
= n-3;Find/(10) 

S"" 1 - 3; Find /(2) 
= x 2 + 4x; Find w( — 5) 
= — 4x + 3; Find w;(6) 
: — |x| + l;Findp(5) 

a + 3;Find£;(-l) 

_2. 4 2x-2. FindA ,(2) 
-2-4 2t+1 + l;Findp(-2) 
:4n + 2;Find/i(n + 2) 

- 3 • 2 a+3 ; Find /i(f) 



x 



l;Find/i(f; 



: t 2 + 1; Find h(t 2 ) 
= 5 n - 1 + l;Find/i( 



392 



10.2 

Functions - Operations on Functions 

Objective: Combine functions using sum, difference, product, quotient 
and composition of functions. 

Several functions can work together in one larger function. There are 5 common 
operations that can be performed on functions. The four basic operations on func- 
tions are adding, subtracting, multiplying, and dividing. The notation for these 
functions is as follows. 

Addition (f + g)(x) = f(x) + g(x) 

Subtraction (/ — g)(x) = f(x) — g(x) 
Multiplication (/ • g)(x) = f(x)g(x) 

Division (l\ x )- f ^ 



gj g{x) 

When we do one of these four basic operations we can simply evaluate the two 
functions at the value and then do the operation with both solutions 

Example 515. 

f(x) =x 2 — x — 2 

g(x) = x + 1 Evaluate / and g at — 3 

find (/+<?)( -3) 

/(-3) = (-3) 2 -(-3)-2 Evaluate / at - 3 
/(-3) = 9 + 3-2 
/(-3) = 10 

g( — 3) = ( — 3) + 1 Evaluate g at — 3 
<?(-3) = -2 

/( — 3) + g( — 3) Add the two functions together 
(10) + (-2) Add 

8 Our Solution 

The process is the same regardless of the operation being performed. 

Example 516. 

h(x) =2x-4 

k[x) = — 3x + l Evaluate h and k at 5 

Find (h-k)(5) 

/i(5) = 2(5)-4 Evaluate h at 5 

393 



ft(5) = 10-4 
h(S) = 6 

k(5) = — 3(5) + 1 Evaluate A; at 5 
fc(5) = — 15 + 1 
fc(5) = -14 

h(5)k(5) Multiply the two results together 
(6)(-14) Multiply 

— 84 Our Solution 

Often as we add, subtract, multiply, or divide functions, we do so in a way that 
keeps the variable. If there is no number to plug into the equations we will simply 
use each equation, in parenthesis, and simplify the expression. 

Example 517. 

f(x) = 2x-4 
g(x) = x 2 — x + 5 Write subtraction problem of functions 
Find (/-s) (X) 

f(x) — g(x) Replace f(x) with (2x — 3) and g(x) with (x 2 — x + 5) 

(2x — 4) — (x 2 — x + 5) Distribute the negative 

2x — 4 — x 2 + x — 5 Combine like terms 

— x 2 + 3x — 9 Our Solution 

The parenthesis are very important when we are replacing f(x) and g(x) with a 
variable. In the previous example we needed the parenthesis to know to distribute 
the negative. 



Example 518. 

f(x) = x 2 — 4x — 5 
g(x) = x — 5 

FindKj(x) 



Write division problem of functions 



m 



Replace f(x) with (x 2 — 4x — 5) and g(x) with (x — 5) 



(x 2 -4x-5) 
(x-5) 



To simplify the fraction we must first factor 



394 



- — -. — — — r — - Divide out common factor of x — 5 

(x — 5) 

x + 1 Our Solution 

Just as we could substitute an expression into evaluating functions, we can substi- 
tute an expression into the operations on functions. 

Example 519. 

f(x) = 2x-l 
g(x) = x + 4 Write as a sum of functions 
Find(f + g)(x 2 ) 

f{x 2 ) + g(x 2 ) Replace x in f(x) and g(x) with x 2 

[2(x 2 ) — 1] + [(x 2 ) + 4] Distribute the + does not change the problem 

2x 2 — 1 + x 2 + 4 Combine like terms 

3x 2 + 3 Our Solution 

Example 520. 

f(x) = 2x-l 

g(x) = x + 4 Write as a product of functions 
Find (f-g) (3x) 

f(3x)g(Sx) Replace x in f(x) and g(x) with 3x 

[2(3x) - 1] [(3x) + 4] Multiply our 2(3x) 

(6a:-l)(3a; + 4) FOIL 

18a; 2 + 24x — 3x — 4 Combine like terms 

18x 2 + 21x-4 Our Solution 

The fifth operation of functions is called composition of functions. A composition 
of functions is a function inside of a function. The notation used for composition 
of functions is: 

(fog)(x) = f(g(x)) 

To calculate a composition of function we will evaluate the inner function and 
substitute the answer into the outer function. This is shown in the following 
example. 



395 



Example 521. 



a(x) = x 2 — 2x + 1 

b(x) = x — 5 Rewrite as a function in function 
Find (a o 6) (3) 

a(6(3)) Evaluate the inner function first, 6(3) 

6(3) = (3) — 5 = — 2 This solution is put into a, a( — 2) 

a(-2) = (-2) 2 -2(-2) + l Evaluate 

o(- 2) =4 + 4+1 Add 

o( — 2) = 9 Our Solution 



396 



We can also evaluate a composition of functions at a variable. In these problems 
we will take the inside function and substitute into the outside function. 



Example 522. 



f(x) = x — X 
g(x) = x + 3 
Find(fog)( x ) 



Rewrite as a function in function 



f(g(x)) Replace g(x) withx + 3 

f(x + 3) Replace the variables in / with (x + 3) 

(x + 3) 2 — (x + 3) Evaluate exponent 

(x 2 + 6x + 9) — (x + 3) Distribute negative 



x 2 + 6x + 9 



x 



x + 5x + 6 



Combine like terms 
Our Solution 



It is important to note that very rarely is (/ o g)(x) the same as (g o f)(x) as the 
following example will show, using the same equations, but compositing them in 
the opposite direction. 



Example 523. 



f(x) = x — X 
g[x) =x + 3 
Find {go f) (x) 



Rewrite as a function in function 



g(f(x)) Replace f(x) with x 2 



g{x 2 
[x 2 — x 

,.2 



X 



X - 



x) Replace the variable in g with (x 2 — x) 
+- 3 Here the parenthesis don't change the expression 
f 3 Our Solution 



World View Note: The term "function" came from Gottfried Wihelm Leibniz, a 
German mathematician from the late 17th century. 



397 



10.2 Practice - Operations on Functions 



Perform the indicated operations. 

1) g( a ) = a 3 + 5a 2 2) f(x) = - 3x 2 + 3x 
/(a) = 2a + 4 g(x)=2x + 5 

Find a(3) + /(3) Find /( - 4) - g( - 4) 

3) g(a) = 3a + 3 4) g(x) =4^ + 3 
/(o) = 2o-2 h(x) = x 3 - 2x 2 

Find(<?+/)(9) Find (g - h)( - 1) 

5) g(x) = x + 3 6) a(x) = - 4x + 1 
/(x) = -a; + 4 /i(a;) = -2x-l 

Find ( o - /) (3) Find o(5) + /i(5) 

7) 5 (x)=x 2 +2 8)^(x) = 3x + l 
/(z) = 2:r + 5 /(x)=x 3 + 3x 2 

Find(a-/)(0) Find g(2) ■ /(2) 

9)tf(t) = t-3 10)/(n) = n-5 

ft (t) = -3t 3 + 6t #(n)=4n + 2 

Find ^(1) + Ml) Find (/ + #)( -8) 

ll)M*) = * + 5 12) a(a) = 3a -2 

5 (t)=3t-5 /i(a)=4a-2 

Find (h-g) (5) Find (a + a) (- 10) 

13) Mn) = 2n - 1 14) ^( x ) = x 2 _ 2 
^(n) = 3n-5 a(x) = 2x + 5 

Find a(0) - #(0) Find ^( _ 6 ) + /,( _ 6 ) 

15) /(a) = - 2a -4 16)a(n)=n 2 -3 
^(a) = a 2 + 3 h(n)=2n-3 

Find (|) (7) Find (a- a) (n) 

17) ( 2; ) = -x 3 -2 18) a(z) = 2x - 3 

a(x)=4x a(s)=x 3 -2x 2 + 2x 

Find (# - a) (x) Find (g - h) (x) 

19) f( x ) = - 3x + 2 20) a(£) = £ - 4 
4( x ) = x 2 + 5x M*) = 2 * 

Find(/-a)(x) Find (g-h)(t) 

21) a(x) = 4x + 5 22) a(i) = - 2i 2 - 5i 
a(x)=x 2 + 5x h(t) = t + h 

Find #(x) • /i(x) Find g(t) ■ h{t) 



398 



23) f(x) 



c — 5x 
g(x) =x + 5 
Find (f + g)(x) 

25) g(n) = n 2 + 5 
f(n) = 3n + 5 
Find g(n) -=- f(n) 

27) g(a) = -2a + 5 
f(a) = 3a + 5 
Find (^)(o) 

29) /i(n) = n 3 + 4n 
g(n) = 4n + 5 
Find h(n) + g(n) 

31) g(n) = n 2 -An 
h(n) =n — 5 



Find g(n 2 ) • /i( 



n 



33) /(x)=2x 

g(:r) = — 3a; — 1 
Find (/+<?)( -4 

35) f(t)=t 2 + U 
g(t) = 4t + 2 

Find f(t 2 ) + g(t 2 ) 

37) g(a)=a 3 + 2a 
h{a) = 3a + A 
Find (£)(-*) 

39) /( n ) = _3 n 2 + 1 
g(n) =2n + 1 
Find (/- 5 )9 

41) f(x) = -4x+l 
g(x) = 4x + 3 
Find (/o ^) (9) 

43) /i(a) = 3a + 3 
^(a) =a+ 1 
Find (hog)(5) 

45) #(x) = 2 + 4 
/i(x) =x 2 — 1 
Find (goh)(10) 



x) 



24) /(x)=4x-4 
g(x)=3x 2 -5 
Find (f + g)(x) 

26) /(x) = 2x + 4 
g(;r) =4x — 5 
Find /(a;) - #(x) 

28) #(£)=i 3 + 3i 2 
/i(t)=3t-5 
Find g(t) - h(t) 

30) /(x)=4x + 2 
g(x) = x 2 + 2x 
Find /(x) -=- g(x) 



32) 



g(n) 
h(n) 
Find 



34) g(a) = 
/z(a) - 
Find 

36) /i(n) 
#(n) = 
Find 

38) <?(*): 

/t(a;) : 
Find 

40) f(n) 
g(n) 
Find 

42) g(x) 
Find 

44) g(t) ~- 
h(t)- 
Find 

46) f(a) 

g(a)-- 

Find 



= n + 5 
= 2n-5 
(g-h)(-3n) 

= -2a 
= 3a 

g{4n) -h h{4n) 

= 3n-2 
= -3n 2 - An 

= -4x + 2 
= x 2 — 5 
g(x 2 ) +h(x 2 ) 

= 3n + 4 
= n 3 — 5n 

= x-l 

(gog)(7) 

= t + 3 
= 2t-5 
(goh)(3) 

= 2a-4 
= a 2 + 2a 
(/°<?)(-4) 



399 



47) /(n) = -4n + 2 
g(n) =n + 4 
Fmd(fog)(9) 

49) a(x) = 2x-4 
h(x) = 2x 3 + 4x 2 
Find (goh)(3) 

51) g(x) = x 2 — 5x 
a(x) = 4x + 4 
Find (go a)(x) 

53) f(a) = -2a + 2 
g(a) =4a 
Find(/oa)(a) 



55) g(x) = 4x 

„3 



4 
/(x) =x 3 — 1 
Find (gof)(x) 

5 



57) g(x) = — x 
/(x) = 2x-3 
Find (gof)(x) 

59) /(*) = 4t + 3 
^(*) = — 4t — 2 

Find(/og)(i) 



48) <?(x) 

/l(x) : 

Find 

50) g(a) : 
Find 

52) 5 (a): 
a(a) - 
Find 

54) g{t) -- 
Find 

56) f(n) 
g(n)-- 
Find 

58) g{t) -- 

/(*) = 
Find 

60) /(x) 
Find 



= 3x + 4 
= x 3 + 3x 
(goh)(3) 

= a 2 + 3 

{gog)(-3) 

= 2a + 4 
= - 4a + 5 
{goh)(a) 

= -t-4 

(gog)(t) 

= -2n 2 - An 
= n + 2 

(f°g)(n) 

= t 3 -t 
= 3t-4 

(gof)(t) 

= 3x-4 

= x 3 + 2x 2 

if°g)(x) 



400 



10.3 

Functions - Inverse Functions 

Objective: Identify and find inverse functions. 

When a value goes into a function it is called the input. The result that we get 
when we evaluate the function is called the output. When working with functions 
sometimes we will know the output and be interested in what input gave us the 
output. To find this we use an inverse function. As the name suggests an inverse 
function undoes whatever the function did. If a function is named /(#), the 
inverse function will be named f~ l (x) (read "/ inverse of x"). The negative one is 
not an exponent, but mearly a symbol to let us know that this function is the 
inverse of /. 

World View Note: The notation used for functions was first introduced by the 
great Swiss mathematician, Leonhard Euler in the 18th century. 

For example, if f(x) = x + 5, we could deduce that the inverse function would be 
f~ 1 (x) = x — 5. If we had an input of 3, we could calculate /(3) = (3) + 5 = 8. Our 
output is 8. If we plug this output into the inverse function we get / _1 (8) = (8) — 
5 = 3, which is the original input. 

Often the functions are much more involved than those described above. It may 
be difficult to determine just by looking at the functions if they are inverses. In 
order to test if two functions, f(x) and g(x) are inverses we will calculate the 
composition of the two functions at x. If / changes the variable x in some way, 
then g undoes whatever / did, then we will be back at x again for our final solu- 
tion. In otherwords, if we simplify (/ o g)(x) the solution will be x. If it is any- 
thing but x the functions are not inverses. 



401 



Example 524. 



„ x^ — 4 

Are f(x) = y3x + 4 and g(x) = — - — inverses? Caculate composition 

%3 ^ 

f(g(x)) Replace g(x) with — - — 

Substitute — - — for variable in f 

i 3 ; 



3 / 3 , :r4 . Divide Qut the g, s 



3 

Vx 3 — 4 + 4 Combine like terms 

v x 3 Take cubed root 

x Simplified to x\ 

Yes, they are inverses! Our Solution 



Example 525. 



x 
Are h(x) = 2x + 5 and g(x) = — — 5 inverses? Calculate composition 



h(g(x)) Replace g(x) with ( tt — 5 j 
/if — — 5 J Substitute ( — — 5 J for variable in h 



2(|-- 5 J + 5 Distrubte2 

x — 10 + 5 Combine like terms 

x — 5 Did not simplify to x 

No, they are not inverses Our Solution 



Example 526. 

Are fix) = and qix) = — inverses? Calculate composition 

J y ' 4x + l ' 3-4x y 

f{g(x)) Replace g(x) with I ^_ 4x 

/( I Substitute I ) for variable in / 

V 3 — Ax J \3 — 4x J 



3 — Ax 

Distribute 3 and 4 into numerators 



\3-4x J 



402 



3x + 6 r. 

|^| Multiply each term by LCD: 3 -Ax 

3-4x "*" 



(3x + 6)(3-4x) 
3-4x 



2(3 -Ax) 



(4z + 8)(3-4:r) 



3-4cc 



f 1(3 -4x) 



Reduce fractions 



3z + 6-2(3 


-Ax) 


Ax + 8 + 1(3 - 


-Ax) 


3a: + 6 — 6 + 8x 



Ax 



3 -Ax 



Distribute 



Combine like terms 



llx 



Divide out 11 



x Simplified to x\ 
Yes, they are inverses Our Solution 



While the composition is useful to show two functions are inverses, a more 
common problem is to find the inverse of a function. If we think of x as our input 
and y as our output from a function, then the inverse will take y as an input and 
give x as the output. This means if we switch x and y in our function we will find 
the inverse! This process is called the switch and solve strategy. 

Switch and solve strategy to find an inverse: 

1. Replace f(x) with y 

2. Switch x and y's 

3. Solve for y 



A. Replace y with / x (s 



Example 527. 



Find the inverse of f(x 



( x ) = (x + 4) 3 - 2 


Replace f(x) with y 


y=(x + A) 3 -2 


Switch x and y 


x = ( 2/ + 4) 3 -2 


Solve for y 


+ 2 +2 


Add 2 to both sides 



403 



x + 2 = (y + A) 3 

Vx+2=y+A 

-A -4 

Vx + 2 



A=y 
f-\x) = Vx + 2-A 



Cube root both sides 
Subtract 4 from both sides 

Replace y with f~ l (x) 
Our Solution 



Example 528. 



Find the inverse of g(x) 



2x-3 



x 



Ax + 2 


2x-3 


4x + 2 


_2y-3 



Ay + 2 



Replace g(x) with y 
Switch x and y 
Multiply by (Ay + 2) 



x(4y + 2) = 2y - 3 Distribute 

Axy + 2x = 2y — 3 Move all y's to one side, rest to other side 
- Axy + 3 — Axy + 3 Subtract 4xy and add 3 to both sides 



2a; + 3 = 2y — Axy Factor out y 



2x + 3 = y(2-Ax) Divide by 2 -Ax 



2 -Ax 2 -Ax 



2x + 3 
2-4x 



?/ Replace y with g 1 (x) 



g 1 (a;) = ; — Our Solution 

v ' 2 -Ax 



In this lesson we looked at two different things, first showing functions are 
inverses by calculating the composition, and second finding an inverse when we 
only have one function. Be careful not to get them backwards. When we already 
have two functions and are asked to show they are inverses, we do not want to use 
the switch and solve strategy, what we want to do is calculate the inverse. There 
may be several ways to represent the same function so the switch and solve 
strategy may not look the way we expect and can lead us to conclude two func- 
tions are not inverses when they are in fact inverses. 



404 



10.3 Practice - Inverse Functions 



State if the given functions are inverses. 



1) 5 ( x ) = _ x 5_3 



3) f(x) 
5) g(x) 

m 

7) /(*) 

/(x)=2x 5 + l 



Find the inverse of each functions. 

(x-2) 5 + 3 

4 
x + 2 

-2x-2 



4- x 



= V-x- 


3 


-x-l 




x-2 
-2x + l 




-x-l 




= -10x + 5 

x — 5 


10 




2 

x + 3 
3x + 2 




x + 2 




_5/*-l 





11) 


/(*) 


13) 


g(x) 


15) 


/(*) 


17) 


/(*) 


19) 


0(aO 


21) 


/(*) 


23) 


#(z) 


25) 


/(*) 


27) 


»(») 


29) 


0(aO 


31) 


g(x) 


33) 


h(x) 


35) 


M 


37) 


m 


39) 


g(x) 



x + 2 
10 -x 



-(x-l) 3 
(x-3) 3 





X 


- 1 




X 


-1 




X 


+ 1 




8 


— 5x 



5x + l 
1+x 3 



4 


-Vax 




2 


X 


+ 1 


X 


+ 2 


_ 7 


— 3x 



-2- 2a; 



a; 



2) <?(x) 
4) /i(x) 

m 

6) /(x) 

/i(x) = 10x + 5 

8) f(x 



x + 2 
x — 5 



10 



5 /a + 1 
' 1 



g(x) = 2x 

8 + 9a; 



10) <?(x 



x-2 
— X 



12) g(x 
14) /(x 

16) #(x 
18) f(x 
20) /(x 

22) g(x 

24) /(x 
26) /i(x 
28) g(x 
30) /(x 
32) f(x 
34) <?(x 
36) /(x 

38) f(x 
40) <?(x 



2 
5a; -9 



Vx+1+2 


-3 


x — 3 


9 + x 


3 


5a; -15 


2 


12 -3a; 


4 


5 /-x + 2 


V 2 


-3- 2a; 


x + 3 


x- 


x + 2 


-x + 2 


3 


5x — 5 


4 


3-2x 5 


(x-l) 3 + 2 


- 1 


x + 1 


3x 
4~ 


-2x + l 



405 



10.4 

Functions - Exponential Functions 

Objective: Solve exponential equations by finding a common base. 

As our study of algebra gets more advanced we begin to study more involved 
functions. One pair of inverse functions we will look at are exponential functions 
and logarithmic functions. Here we will look at exponential functions and then we 
will consider logarithmic functions in another lesson. Exponential functions are 
functions where the variable is in the exponent such as /(#) = a x . (It is important 
not to confuse exponential functions with polynomial functions where the variable 
is in the base such as f(x) =x 2 ). 

World View Note One common application of exponential functions is popula- 
tion growth. According to the 2009 CIA World Factbook, the country with the 
highest population growth rate is a tie between the United Arab Emirates (north 
of Saudi Arabia) and Burundi (central Africa) at 3.69%. There are 32 countries 
with negative growth rates, the lowest being the Northern Mariana Islands (north 
of Australia) at - 7.08%. 

Solving exponetial equations cannot be done using the skill set we have seen in 
the past. For example, if 3 X = 9, we cannot take the x — root of 9 because we do 
not know what the index is and this doesn't get us any closer to finding x. How- 
ever, we may notice that 9 is 3 2 . We can then conclude that if 3 X = 3 2 then x = 2. 
This is the process we will use to solve exponential functions. If we can re-write a 
problem so the bases match, then the exponents must also match. 



Example 529. 








5 2x+1 = 125 


Rewrite 125 as 5 3 




g2a; + l_ g3 


Same base, set exponents equal 




2x + 1 = 3 


Solve 




-1-1 


Subtract 1 from both sides 




2x = 2 


Divide both sides by 2 




~2~ ~2~ 






x= 1 


Our Solution 



Sometimes we may have to do work on both sides of the equation to get a 
common base. As we do so, we will use various exponent properties to help. First 
we will use the exponent property that states (a x ) y = a xy . 



406 



Example 530. 






8 3a; = 32 




(2 3 ) 3:E = 2 5 




2 9x = 2 5 




9x = 5 




"9" ~9~ 




5 
X= 9 



Rewrite 8 as 2 3 and 32 as 2 5 

Multiply exponents 3 and 3a; 

Same base, set exponents equal 

Solve 

Divide both sides by 9 

Our Solution 

As we multiply exponents we may need to distribute if there are several terms 
involved. 



Rewrite 27 as 3 3 and 81 as 3 4 (9 2 would not be same base) 

Multiply exponents 3 (3a; + 5) and 4(4x + 1) 

Same base, set exponents equal 

Move variables to one side 

Subtract 9a; from both sides 

Subtract 4 from both sides 

Divide both sides by 7 

Our Solution 

Another useful exponent property is that negative exponents will give us a recip- 
rocal, — = a~ n 



Example 531 


• 


9'73a:+5 _ 


_ 01 4a: +1 


/g3\3x+5_ 


/o4N4x+l 


o9x + 15 


_ o16ie+4 


9a; + 15 = 


: 16a; + 4 


-9a; 


-9s 


15: 


= 7x + 4 


-4 


-4 




11 = 7a; 




T~ ~T 




11 
T = x 



Example 532. 



l\ 2x _ 1 

— I = 3 7x x Rewrite — as 3 2 (negative exponet to flip) 

9 J 9 

(3" 2 ) 2:E = S 7 "- 1 Multiply exponents - 2 and 2a; 

3~ Ax = 3 7x ~ l Same base, set exponets equal 

— 4x = 7x — 1 Subtract 7x from both sides 

— 7a; — 7a; 



11a; = — 1 Divide by — 11 



11 -11 

IT 



x = — Our Solution 



If we have several factors with the same base on one side of the equation we can 
add the exponents using the property that states a x a v = a x+v . 



407 



Example 533. 

5 4x ■ 5 2x ~ 1 = 5 3:E+11 Add exponents on left, combing like terms 

562-1 _ 532+11 Same base, set exponents equal 

62 — 1 = 3x + 1 1 Move variables to one sides 

— 3a; — 3x Subtract 3x from both sides 

3x — 1 = 11 Add 1 to both sides 
+ 1 +1 

3rr = 1 2 Divide both sides by 3 
"3" ~3~ 

x = 4 Our Solution 

It may take a bit of practice to get use to knowing which base to use, but as we 
practice we will get much quicker at knowing which base to use. As we do so, we 
will use our exponent properties to help us simplify. Again, below are the proper- 
ties we used to simplify. 

(a x ) y = a xy and — = a~ n and a x a y = a x+y 

We could see all three properties used in the same problem as we get a common 
base. This is shown in the next example. 

Example 534. 

(-. \ 32+1 / 1 \ x+3 

— I = 32 • [ — I Write with a common base of 2 

{2 4 ) 2x - b ■ {2~ 2 f x+1 = 2 5 • (2" 1 ) a;+3 Multiply exponents, distributing as needed 

282-20. 2-62-2 _ 25 . 2~ x ~ 3 Add exponents, combining like terms 

222-22 _ 2-2+2 Same base, set exponents equal 

2x — 22 = — x + 2 Move variables to one side 

+ x + x Add x to both sides 

3x-22 = 2 Add 22 to both sides 
+ 22 + 22 

3x = 24 Divide both sides by 3 
~3~ ~3~ 

x = 8 Our Solution 

All the problems we have solved here we were able to write with a common base. 
However, not all problems can be written with a common base, for example, 2 = 
1(F, we cannot write this problem with a common base. To solve problems like 
this we will need to use the inverse of an exponential function. The inverse is 
called a logarithmic function, which we will discuss in another secion. 



408 



10.4 Practice - Exponential Functions 



Solve each equation. 

I) 3 1 - 2n = 3 1 " 3 " 
3) 4 2a =l 

5) (^)- fc = 125" 2fc - 2 
7) 6 2m+1 = — 

' 36 

9) 6" 3:c = 36 

II) 64 6 = 2 5 
13) (i)* =16 

15) 4 3a = 4 3 

17) 36 3x = 216 2a;+1 

19) 9 2n+3 = 243 

o3a; — 2 o3ie+1 

3-2^ = 33 
cm+2 _ er — m 

(i) 6 " 1 = 216 

6 2-2x = g2 



21 

23 
25 
27 
29 
31 

33 



A . O — 3n — 1 _ }_ 
^3k-3 . ^2-2fc_ jg-/ 



35) 9" 2a; -(^) 3 ^ = 243- a; 
37) 64 n - 2 -16 n+2 =(|) 3 ™- 1 
39) 5" 3 "- 3 -5 2n =l 



2) 4 2x = — 

I 16 

4) I6" 3p = 64- 3p 

6) 625 

8) 6 2r - 3 = 6 r - 3 



n-2_ _J_ 
125 



10 
12 

14 
16 
18 
20 
22 
24 
26 
28 
30 
32 
34 
36 
38 
40 



5 2n = 5 -n 

216" 3t ' = 36 31 ' 
27 _ 2 „-i = 9 

4-^ = 64 
64 x+ 2=16 

16 2fc = - 

64 

243 p = 27" 3p 

A2n A2 — 2>n 

625 2x = 25 
216 2n = 36 



1 \3t>-2. 



(x) 



64 



l-v 



216 _ R 3a 



32 2 P -2. 8 p = (l)2p 



o2ra _ o3m 1 

3 2-x. 3 3m =1 



43r. 4-3r _ J_ 
64 



409 



10.5 

Functions - Logarithmic Functions 

Objective: Convert between logarithms and exponents and use that 
relationship to solve basic logarithmic equations. 

The inverse of an exponential function is a new function known as a logarithm. 
Lograithms are studied in detail in advanced algebra, here we will take an intro- 
ductory look at how logarithms works. When working with radicals we found that 
there were two ways to write radicals. The expression vV could be written as 
a™. Each form has its advantages, thus we need to be comfortable using both the 
radical form and the rational exponent form. Similarly an exponent can be 
written in two forms, each with its own advantages. The first form we are very 
familiar with, b x = a, where b is the base, a can be thought of as our answer, and 
x is the exponent. The second way to write this is with a logarithm, log&a = x. 
The word "log" tells us that we are in this new form. The variables all still mean 
the same thing, b is still the base, a can still be thought of as our answer. 

Using this idea the problem 5 2 = 25 could also be written as logs25 = 2. Both 
mean the same thing, both are still the same exponent problem, but just as roots 
can be written in radical form or rational exponent form, both our forms have 
their own advantages. The most important thing to be comfortable doing with 
logarithms and exponents is to be able to switch back and forth between the two 
forms. This is what is shown in the next few examples. 

Example 535. 

Write each exponential equation in logarithmic form 

m 3 = 5 Identify base, m , answer, 5, and exponent 3 
log m 5 = 3 Our Solution 

7 2 = b Identify base, 7, answer, b, and exponent, 2 
log7& = 2 Our Solution 

2 Y 16 ja *•* v 2 16 , 

— I = — — identity base, — , answer, — , and exponent 4 

16 
log2 — = 4 Our Solution 



3 



81 



Example 536. 

Write each logarithmic equation in exponential form 

logztl6 = 2 Identify base, 4, answer, 16, and exponent, 2 
4 2 = 16 Our Solution 



410 



log3X = 7 Identify base, 3, answer, x, and exponent, 7 
S 7 = x Our Solution 



logg3 = — Identify base, 9, answer, 3, and exponent, — 



9 2 = 3 Our Solution 



Once we are comfortable switching between logarithmic and exponential form we 
are able to evaluate and solve logarithmic expressions and equations. We will first 
evaluate logarithmic expressions. An easy way to evaluate a logarithm is to set 
the logarithm equal to x and change it into an exponential equation. 



Example 537. 



Evaluate log 2 64 

log 2 64 = x 

T = 64 

2 X = 2 6 

x = 6 



Set logarithm equal to x 
Change to exponent form 
Write as common base, 64 = 2 6 
Same base, set exponents equal 
Our Solution 



Example 538. 



Evaluate logi2s5 

logi 25 5 = x 

125* = 5 

(5 3 ) x = 5 

5 3a; = 5 

3a; =1 

1 
X= 3 



Example 539. 



Set logarithm equal to x 

Change to exponent form 

Write as common base, 125 = 5 3 

Multiply exponents 

Same base, set exponents equal (5 = 5 1 ) 

Solve 

Divide both sides by 3 

Our Solution 



log 3 



Evaluate log3- 

J^ 

'27 

S x = 

2 

3 X = 3" 3 

x = — 3 



'27 

= x 

l_ 

27 



Set logarithm equal to x 
Change to exponent form 

Write as common base, — — = 3 -3 

27 

Same base, set exponents equal 
Our Solution 



World View Note: Dutch mathematician Adriaan Vlacq published a text in 
1628 which listed logarithms calculated out from 1 to 100,000! 



411 



Solve equations with logarithms is done in a very similar way, we simply will 
change the equation into exponential form and try to solve the resulting equation. 



Example 540. 



log5X = 2 Change to exponential form 
5 2 = x Evaluate exponent 
25 = x Our Solution 



Example 541. 








log 2 (3:r + 5)=4 


Change to exponential form 




2 4 = 3x + 5 


Evaluate exponent 




16 = 3x + 5 


Solve 




-5 -5 


Subtract 5 from both sides 




11 = 3a; 


Divide both sides by 3 




~3~ ~3~ 






11 


Our Solution 



Example 542. 








log x 8 = 3 


Change to exponential form 




x 3 = 8 


Cube root of both sides 




x = 2 


Our Solution 



There is one base on a logarithm that gets used more often than any other base, 
base 10. Similar to square roots not writting the common index of 2 in the rad- 
ical, we don't write the common base of 10 in the logarithm. So if we are working 
on a problem with no base written we will always assume that base is base 10. 

Example 543. 



logx = 

10" 2 : 
1 



100 



- 2 Rewrite as exponent, 10 is base 

: x Evaluate, remember negative exponent is fraction 

: x Our Solution 



This lesson has introduced the idea of logarithms, changing between logs and 
exponents, evaluating logarithms, and solving basic logarithmic equations. In an 
advanced algebra course logarithms will be studied in much greater detail. 



412 



10.5 Practice - Logarithmic Functions 

Rewrite each equation in exponential form. 

l)log 9 81 = 2 2)log b a=-16 

3)log 7 ^ = -2 4)log 16 256 = 2 

5)logi 3 169 = 2 6)logul = 

Rewrite each equations in logarithmic form. 

7)8°=1 8 )i7-2 = J_ 

' I 289 

9)152 = 225 10) 144* =12 

11) 64^ = 2 12) 19 2 = 361 

Evaluate each expression. 

13) log 125 5 14) logs 125 

15)log 3 43y 16)log 7 l 

17) log 4 16 18) log 4 ^ 

19) log 6 36 20) log 36 6 

21) log 2 64 22) logs 243 



Solve each equation. 

23) logs x=l 24) log 8 k = 3 

25) log 2 x = - 2 26) log n = 3 

27)lognA; = 2 28) log 4 p = 4 

29) log 9 (n + 9) =4 30) logn (x - 4) = - 1 

31) logs (-3m) = 3 32)log 2 -8r = l 

33) logn (a: + 5) = - 1 34)log 7 -3n = 4 

35) log 4 (66 + 4) = 36) logn (10u + 1) = - 1 

37) logs ( - 10x + 4) = 4 38) log 9 (7 - 6x) = - 2 

39) log 2 (10-5a)=3 40) log 8 (3fc - 1) = 1 



413 



10.6 

Functions - Compound Interest 



Objective: Calculate final account balances using the formulas for com- 
pound and continuous interest. 

An application of exponential functions is compound interest. When money is 
invested in an account (or given out on loan) a certain amount is added to the 
balance. This money added to the balance is called interest. Once that interest is 
added to the balance, it will earn more interest during the next compounding 
period. This idea of earning interest on interest is called compound interest. For 
example, if you invest $100 at 10% interest compounded annually, after one year 
you will earn $10 in interest, giving you a new balance of $110. The next year 
you will earn another 10% or $11, giving you a new balance of $121. The third 
year you will earn another 10% or $12.10, giving you a new balance of $133.10. 
This pattern will continue each year until you close the account. 

There are several ways interest can be paid. The first way, as described above, is 
compounded annually. In this model the interest is paid once per year. But 
interest can be compounded more often. Some common compounds include com- 
pounded semi-annually (twice per year), quarterly (four times per year, such as 
quarterly taxes), monthly (12 times per year, such as a savings account), weekly 
(52 times per year), or even daily (365 times per year, such as some student 
loans). When interest is compounded in any of these ways we can calculate the 
balance after any amount of time using the following formula: 



Compound Interest Formula: A = P( 1 + 



nt 



n 

A = Final Amount 



P = Principle (starting balance) 

r = Interest rate (as a decimal) 

n = number of compounds per year 

t = time (in years) 



Example 544. 



If you take a car loan for $25000 with an interest rate of 6.5% compounded quar- 
terly, no payments required for the first five years, what will your balance be at 
the end of those five years? 

P = 25000, r = 0.065, n = 4, t = 5 Identify each variable 

(0 Ofi^ \ 4 ' 5 
1 H — — — I Plug each value into formula, evaluate parenthesis 

A = 25000(1. 01625) 4 ' 5 Multiply exponents 



414 



A = 25000(1.01625) 20 Evaluate exponent 

A = 25000(1.38041977...) Multiply 
,4 = 34510.49 

$34,510.49 Our Solution 



We can also find a missing part of the equation by using our techniques for 
solving equations. 



Example 545. 

What principle will amount to $3000 if invested at 6.5% compounded weekly for 
4 years? 

A = 3000, r = 0.065, n = 52, t = 4 Identify each variable 

(n nfi^i \ 52 ' 4 
1 -\ — '— - — I Evaluate parentheses 

52 ) 

3000 = P(1.00125) 52 ' 4 Multiply exponent 
3000 = P(1.00125) 208 Evaluate exponent 
3000 = P(1.296719528...) Divide each side by 1.296719528... 



1.296719528... 1.296719528... 

2313.53 = P Solution for P 
$2313.53 Our Solution 



It is interesting to compare equal investments that are made at several different 
types of compounds. The next few examples do just that. 



Example 546. 

If $4000 is invested in an account paying 3% interest compounded monthly, what 
is the balance after 7 years? 



P = 4000, r = 0.03, n=12,t = 7 Identify each variable 

/ 0.03 V 2 ' 7 

A = 4000 1 1 + — I Plug each value into formula, evaluate parentheses 

A = 4000(1. 0025) 12 ' 7 Multiply exponents 

A = 4000(1. 0025) 84 Evaluate exponent 

A = 4000(1.2333548) Multiply 
A = 4933.42 

$4933.42 Our Solution 



415 



To investigate what happens to the balance if the compounds happen more often, 
we will consider the same problem, this time with interest compounded daily. 



Example 547. 

If $4000 is invested in an account paying 
the balance after 7 years? 



Vo interest compounded daily, what is 



P = 4000, r = 0.03, n = 365, t= 7 Identify each variable 

A = 4000(1.00008219.. .) 365 ' 7 

A = 4000(1. 00008219.. .) 2555 

4 = 4000(1.23366741....) 

4 = 4934.67 

$4934.67 Our Solution 



Plug each value into formula, evaluate parenthesis 

Multiply exponent 
Evaluate exponent 
Multiply 



While this difference is not very large, it is a bit higher. The table below shows 
the result for the same problem with different compounds. 



Compound 


Balance 


Annually 


$4919.50 


Semi- Annually 


$4927.02 


Quarterly 


$4930.85 


Monthly 


$4933.42 


Weekly 


$4934.41 


Daily 


$4934.67 



As the table illustrates, the more often interest is compounded, the higher the 
final balance will be. The reason is, because we are calculating compound interest 
or interest on interest. So once interest is paid into the account it will start 
earning interest for the next compound and thus giving a higher final balance. 
The next question one might consider is what is the maximum number of com- 
pounds possible? We actually have a way to calculate interest compounded an 
infinite number of times a year. This is when the interest is compounded continu- 
ously. When we see the word "continuously" we will know that we cannot use the 
first formula. Instead we will use the following formula: 

Interest Compounded Continuously: A = Pe vt 

A = Final Amount 

P = Principle (starting balance) 

e = a constant approximately 2.71828183. . . . 

r = Interest rate (written as a decimal) 

t = time (years) 



416 



The variable e is a constant similar in idea to pi (jr) in that it goes on forever 
without repeat or pattern, but just as pi (it) naturally occurs in several geometry 
applications, so does e appear in many exponential applications, continuous 
interest being one of them. If you have a scientific calculator you probably have 
an e button (often using the 2nd or shift key, then hit In) that will be useful in 
calculating interest compounded continuously. 

World View Note: e first appeared in 1618 in Scottish mathematician's 
Napier's work on logarithms. However it was Euler in Switzerland who used the 
letter e first to represent this value. Some say he used e because his name begins 
with E. Others, say it is because exponent starts with e. Others say it is because 
Euler's work already had the letter a in use, so e would be the next value. What- 
ever the reason, ever since he used it in 1731, e became the natural base. 

Example 548. 

If $4000 is invested in an account paying 3% interest compounded continuously, 
what is the balance after 7 years? 

P = 4000, r = 0.03, t = 7 Identify each of the variables 

A = 4000e a03 ' 7 Multiply exponent 

A = 4000e a21 Evaluate e a21 

A = 4000(1.23367806...) Multiply 
,4 = 4934.71 

$4934.71 Our Solution 

Albert Einstein once said that the most powerful force in the universe is com- 
pound interest. Consider the following example, illustrating how powerful com- 
pound interest can be. 

Example 549. 

If you invest $6.16 in an account paying 12% interest compounded continuously 
for 100 years, and that is all you have to leave your children as an inheritance, 
what will the final balance be that they will receive? 

P = 6.16, r = 0.12, t = 100 Identify each of the variables 

A = 6.16e ai2 - 100 Multiply exponent 

A = 6.16e 12 Evaluate 

A = 6.16(162, 544.79) Multiply 
A = 1,002, 569.52 

$1,002,569.52 Our Solution 

In 100 years that one time investment of $6.16 investment grew to over one mil- 
lion dollars! That's the power of compound interest! 



417 



10.6 Practice - Compound Interest 



Solve 

1) Find each of the following: 

a. $500 invested at 4% compounded annually for 10 years. 

b. $600 invested at 6% compounded annually for 6 years. 

c. $750 invested at 3% compounded annually for 8 years. 

d. $1500 invested at 4% compounded semiannually for 7 years. 

e. $900 invested at 6% compounded semiannually for 5 years. 

f. $950 invested at 4% compounded semiannually for 12 years. 

g. $2000 invested at 5% compounded quarterly for 6 years. 
h. $2250 invested at 4% compounded quarterly for 9 years. 
i. $3500 invested at 6% compounded quarterly for 12 years. 



418 



j. All of the above compounded continuously. 

2) What principal will amount to $2000 if invested at 4% interest compounded 

semiannually for 5 years? 

3) What principal will amount to $3500 if invested at 4% interest compounded 

quarterly for 5 years? 

4) What principal will amount to $3000 if invested at 3% interest compounded 

semiannually for 10 years? 

5) What principal will amount to $2500 if invested at 5% interest compounded 

semiannually for 7.5 years? 

6) What principal will amount to $1750 if invested at 3% interest compounded 

quarterly for 5 years? 

7) A thousand dollars is left in a bank savings account drawing 7% interest, 

compounded quarterly for 10 years. What is the balance at the end of that 
time? 

8) A thousand dollars is left in a credit union drawing 7% compounded monthly. 

What is the balance at the end of 10 years? 

9) $1750 is invested in an account earning 13.5% interest compounded monthly 

for a 2 year period. What is the balance at the end of 9 years? 

10) You lend out $5500 at 10% compounded monthly. If the debt is repaid in 18 

months, what is the total owed at the time of repayment? 

11) A $10,000 Treasury Bill earned 16% compounded monthly. If the bill 

matured in 2 years, what was it worth at maturity? 

12) You borrow $25000 at 12.25% interest compounded monthly. If you are 
unable to make any payments the first year, how much do you owe, excluding 
penalties? 

13) A savings institution advertises 7% annual interest, compounded daily, How 
much more interest would you earn over the bank savings account or credit 
union in problems 7 and 8? 

14) An 8.5% account earns continuous interest. If $2500 is deposited for 5 years, 
what is the total accumulated? 

15) You lend $100 at 10% continuous interest. If you are repaid 2 months later, 
what is owed? 



419 



10.7 



Functions - Trigonometric Functions 




Hypotenuse 



A 



Objective: Solve for a missing side of a right triangle using trigono- 
metric ratios. 

There are six special functions that describe the relationship between the sides of 
a right triangle and the angles of the triangle. We will discuss three of the func- 
tions here. The three functions are called the sine, cosine, and tangent (the three 
others are cosecant, secant, and cotangent, but we will not need to use them 
here). 

To the right is a picture of a right tri- 
angle. Based on which angle we are 
interested in on a given problem we 
will name the three sides in relation- 
ship to that angle. In the picture, Opposite 
angle A is the angle we will use to 
name the other sides. The longest side, 
the side opposite the right angle is 

always called the hypotenouse. The Adjacent 

side across from the angle A is called 
the opposite side. 

The third side, the side between our angle and the right angle is called the adja- 
cent side. It is important to remember that the opposite and adjacent sides are 
named in relationship to the angle A or the angle we are using in a problem. If 
the angle had been the top angle, the opposite and adjacent sides would have 
been switched. 

The three trigonometric funtions are functions taken of angles. When an angle 
goes into the function, the output is a ratio of two of the triangle sides. The 
ratios are as describe below: 

sin0 = u ° PP0Site cosfl = u adjaCent tanfl = PP osite 

hypotenuse hypotenuse adjacent 

The "weird" variable 9 is a greek letter, pronounced "theta" and is close in idea to 
our letter "t". Often working with triangles, the angles are repesented with Greek 
letters, in honor of the Ancient Greeks who developed much of Geometry. Some 
students remember the three ratios by remembering the word "SOH CAH TOA" 
where each letter is the first word of: "Sine: Opposite over Hypotenuse; Cosine: 
Adjacent over Hypotenuse; and Tangent: Opposite over Adjacent." Knowing how 
to use each of these relationships is fundamental to solving problems using 
trigonometry. 

World View Note: The word "sine" comes from a mistranslation of the Arab 
word jayb 

Example 550. 



420 



Using the diagram at right, find each 
of the following: sin^, cos#, tan#, sina, 
cosa, and tana. 

First we will find the three ratios of 9. 
The hypotenuse is 10, from 9, the 
opposite side is 6 and the adjacent 
side is 8. So we fill in the following: 



sin#: 



cos^ : 



tan#: 



opposite 6 

hypotenuse 10 

adjacent 8 

hypotenuse 10 



3 

5 

4 

' 5 



opposite 
adjacent 



6 



Now we will find the three ratios of a. 
The hypotenuse is 10, from a, the 
opposite side is 8 and the adjacent 
side is 6. So we fill in the following: 




Adjacent of 9 
Opposite of a 



Opposite of 9 
Adjacent of a 




10 
Hypotenuse 



sma : 



cosa : 



tana: 



opposite 8 4 

hypotenuse 10 5 

adjacent 6 3 

hypotenuse 10 5 

opposite 8 4 

adjacent 6 3 



We can either use a trigonometry table or a calculator to find decimal values for 
sine, cosine, or tangent of any angle. We only put angle values into the trigono- 
metric functions, never values for sides. Using either a table or a calculator, we 
can solve the next example. 

Example 551. 

sin 42° Use calculator or table 
0.669 Our Solution 

tan 12° Use calculator or table 
0.213 Our Solution 

cos 18° Use calculator or table 
0.951 Our Solution 

By combining the ratios together with the decimal approximations the calculator 
or table gives us, we can solve for missing sides of a triangle. The trick will be to 
determine which angle we are working with, naming the sides we are working 
with, and deciding which trig function can be used with the sides we have. 



421 



Example 552. 

Find the measure of the missing side. 

x 




We will be using the angle marked 
25°, from this angle, the side marked 4 
is the opposite side and the side 
marked x is the adjacent side. 

The trig ratio that uses the opposite 
and adjacent sides is tangent. So we 
will take the tangent of our angle. 



tan25° = — Tangent is opposite over adjacent 

x 

0.466 4 

— - — = — Evaluate tan25°, put over 1 so we have proportion 
1 x 



0.466a; = 4 Find cross product 
0A66 0T466 Divide both sides by 0.466 
x = 8.58 Our Solution 



Example 553. 

Find the measure of the missing side. 



x 




We will be using the angle marked 
70°. From this angle, the x is the 
adjacent side and the 9 is the 
hypotenuse. 

The trig ratio that uses adjacent and 
hypotenuse is the cosine. So we will 
take the cosine of our angle. 



x 



cos70° = — Cosine is adjacent over hypotenuse 



0.342 x 



1 



9 



Evaluate cos70°, put over 1 so we have a proportion 



3.08 = lx Find the cross product. 
3.08 = x Our Solution. 



422 



10.7 Practice - Trigonometric Functions 



Find the value of each. Round your answers to the nearest 
ten-thousandth. 



1) cos 71° 
3) sin 75° 



2) cos 23° 
4) sin 50° 



Find the value of the trig function indicated. 

5) sin 9 6) tan 9 




sin 




15 



8 



9 
17 



9) sin 9 



8\/2 



7) sin 9 




10) cos 9 




20 



Find the measure of each side indicated. Round to the nearest tenth. 



11 



12) 




A 



B 




423 



13) 



14) 



B 




24X A 



B 

5 

C 




A 



15) 



16) 





17) 



C 




18) 



C 




B 



19) 



20) 



B 



6 c 



\71 



x 



B 




424 



2V 



22) 



A 




C 



x 



B 



287 12 



23) 



B 



x 



/67T 



24) 



B 




C ~x~ A 



25) 



B 



$7.2* 
A 4 C 



x 



26) 



A 
7 

cti 



16^^ B 



x 



27) 



B 




4 C 



28) 



A 



B -64° 



13.1 



C 



425 



29) 



30) 





B 



31: 



B 




A 



32) 



B 




33) 



34) 





35) 




36) 



B 




A 3 c 



426 



37) 



38) 



C 





39) 



40) 



C 




B 



C 




427 



10.8 

Functions - Inverse Trigonometric Functions 



Objective: Solve for missing angles of a right triangle using inverse 
trigonometry. 

We used a special function, one of the trig functions, to take an angle of a triangle 
and find the side length. Here we will do the opposite, take the side lengths and 
find the angle. Because this is the opposite operation, we will use the inverse 
function of each of the trig ratios we saw before. The notation we will use for the 
inverse trig functions will be similar to the inverse notation we used with func- 
tions. 

^ opposite \ cos -if adjacent \ $ t J opposite \ = Q 

\ hypotenuse J \ hypotenuse J \ adjacent J 

Just as with inverse functions, the — 1 is not an exponent, it is a notation to tell 
us that these are inverse functions. While the regular trig functions take angles as 
inputs, these inverse functions will always take a ratio of sides as inputs. We can 
calculate inverse trig values using a table or a calculator (usually pressing shift or 
2nd first). 

Example 554. 

sinA = 0.5 We don't know the angle so we use an inverse trig function 
sin _1 (0.5) = A Evaluate using table or calculator 
30° = A Our Solution 



cosB = 0.667 We don't know the angle so we use an inverse trig function 
cos _1 (0.667) = B Evaluate using table or calculator 
48° = B Our Solution 

tanC = 1.54 We don't know the angle so we use an inverse trig function 
tan -1 (1.54) = C Evaluate using table or calculator 
57° = C Our Solution 

If we have two sides of a triangle, we can easily calculate their ratio as a decimal 
and then use one of the inverse trig functions to find a missing angle. 

Example 555. 

Find the indicated angle. 

428 




sin 



12 



17 

sin-^OJOG) 
45° 



From angle 9 the given sides are the 
opposite (12) and the hypotenuse (17). 



The trig function that uses opposite 
and hypotenuse is the sine 

Because we are looking for an angle 
we use the inverse sine 



Sine is opposite over hyptenuse, use inverse to find angle 

Evaluate fraction, take sine inverse using table or calculator 
Our Solution 



Example 556. 

Find the indicated angle 




From the angle a, the given sides are 
the opposite (5) and the adjacent (3) 

The trig function that uses opposite 
and adjacent is the tangent 

As we are looking for an angle we will 
use the inverse tangent. 



tan '■( — ) Tangent is opposite over adjacent. Use inverse to find angle 

tan _1 (1.667) Evaluate fraction, take tangent inverse on table or calculator 
59° Our Solution 

Using a combination of trig functions and inverse trig functions, if we are given 
two parts of a right triangle (two sides or a side and an angle), we can find all the 
other sides and angles of the triangle. This is called solving a triangle. 

When we are solving a triangle, we can use trig ratios to solve for all the missing 
parts of it, but there are some properties from geometry that may be helpful 
along the way. 

The angles of a triangle always add up to 180°, because we have a right triangle, 
90° are used up in the right angle, that means there are another 90° left in the 
two acute angles. In other words, the smaller two angles will always add to 90, if 
we know one angle, we can quickly find the other by subtracting from 90. 

Another trick is on the sides of the angles. If we know two sides of the right tri- 
angle, we can use the Pythagorean Theorem to find the third side. The 
Pythagorean Theorem states that if c is the hypotenuse of the triangle, and a and 



429 



b are the other two sides (legs), then we can use the following formula, a 2 + b 2 = c 2 
to find a missing side. 

Often when solving triangles we use trigonometry to find one part, then use the 
angle sum and/or the Pythagorean Theorem to find the other two parts. 



Example 557. 

Solve the triangle 




We have one angle and one side. We 
can use these to find either other side. 
We will find the other leg, the adja- 
cent side to the 35° angle. 

The 5 is the opposite side, so we will 
use the tangent to find the leg. 



tan35° = - 
0.700 5 



[ x 

0.700x= 5 
07T00 07T00 

x = 7.1 



Tangent is opposite over adjacent 

Evaluate tangent, put it over one so we have a proportion 
Find cross product 
Divide both sides by 0.700 
The missing leg. 



a 2 + b 2 = 

5 2 + 7.1 2 = 

25 + 50.41 = 

75.41 = 

8.7: 



We can now use pythagorean thorem to find hypotenuse, c 

Evaluate exponents 

Add 

Square root both sides 

The hypotenuse 



90° — 35° To find the missing angle we subtract from 90° 
55° The missing angle 




Our Solution 



In the previous example, once we found the leg to be 7.1 we could have used the 
sine function on the 35° angle to get the hypotenuse and then any inverse trig 



430 



function to find the missing angle and we would have found the same answers. 
The angle sum and pythagorean theorem are just nice shortcuts to solve the 
problem quicker. 



Example 558. 
Solve the triangle 




In this triangle we have two sides. We 
will first find the angle on the right 
side, adjacent to 3 and opposite from 
the 9. 

Tangent uses opposite and adjacent 

To find an angle we use the inverse 
tangent. 



9\ 

tan i | — ) Evaluate fraction 

w 

tan _1 (3) Evaluate tangent 

71.6° The angle on the right side 

90° — 71.6° Subtract angle from 90° to get other angle 

18.4° The angle on the left side 



a 2 + b 2 -- 

9 2 + 3 2 = 

81 + 9 = 

90 = 

3\/l0or9.5 



c 2 Pythagorean theorem to find hypotenuse 

c 2 E valuat e exp onent s 

c 2 Add 

c 2 Square root both sides 

= c The hypotenuse 



9.5 




Our Solution 



World View Note: Ancient Babylonian astronomers kept detailed records of the 
starts, planets, and eclipses using trigonometric ratios as early as 1900 BC! 



431 



10.8 Practice - Inverse Trigonometric Functions 

Find each angle measure to the nearest degree. 

1) sin Z = 0.4848 2) sin Y = 0.6293 

3) sin Y = 0.6561 4) cos Y = 0.6157 



Find the measure of the indicated angle to the nearest degree. 

5) 6) 





7) 



30 




8) 




9) 




10) 




432 



Find the measure of each angle indicated. Round to the nearest tenth. 

11) 12) 

B A B 




C 




A 



13) 



A 




14) 



C 




15) 



B 




A 4 C 



16) 



B 




17) 




B 



18) 



B 



5.6 



C 




15.3 



A 



433 



19) 



C 



20) 





2V 



22) 



B 



C 



£l\ A 



B 




23) 



C 




24) 



A 



C 



B 



25) 



C 




15.7 



A 



26) 



B 




434 



27) 



A 




B 



28) 



C 



B 




29) 



30) 



C 




B 




A 



Solve each triangle. Round answers to the nearest tenth. 



31: 




32) 



B 




33) 



34) 



C 




A 



B 



A 



C 




B 



435 



35) 



B 



3 
C 




A 



37) 



B 




C 



36) 



A 



C 



21° 



B 



38) 




B 



39) 



B 




C 



40) 



B 



C 



14 



6.8 



A 



436 



437 



Answers - Chapter 



o.i 

1) -2 

2) 5 

3) 2 

4) 2 

5) -6 

6) -5 

7) 8 

8) 

9) -2 

10) -5 

11) 4 

12) -7 

13) 3 

14) -9 

15) -2 

16) -9 

17) -1 

18) -2 

19) -3 

20) 2 

21) -7 



Answers - Integers 


22) 


43) -20 


23) 11 


44) 27 


24) 9 


45) -24 


25) -3 




26) -4 


46) -3 


27) -3 


47) 7 


28) 4 


48) 3 


29) 


49) 2 


30) -8 


50) 5 


31) -4 


51) 2 


32) -35 




33) -80 


52) 9 


34) 14 


53) 7 


35) 8 


54) -1( 


36) 6 


55) 4 


37) -56 


56) 10 


38) -6 


57) -8 


39) -36 




40) 63 


58) 6 


41) -10 


59) -6 


42) 4 


60) -9 



0.2 



2) 
3 ) 5 



Answers - Fractions 

5) 
6 ), 



438 



7) 
8) 

9) 
10 
11 
12 
13 
14 
15 
16 
17 
18 
19 

20 
21 

22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 



4 
9 

2 
3 

13 

T 



3 

4 

33 

20 

33 

56 



18 

T 

i 

2 



19 

20 



33 

34 

35 
36 
37 

38 
39 
40 
41 

42 

43 

44 

45 

46 

47 

48 

49 

50 

51 
52 
53 

54 

55 
56 

57 
58 



17 
T~5 

7 

To 



5 

14 



20 
21 

2 
9 

4 
3 



21 
26 



25 
21 



3 

2 

5 

27 



40 
~9~ 



1 
10 

45 
T 



13 
15 

4 
27 

32 
65 

1 
15 



10 

T 



31 

~8~ 



59 

60 

61 

62 

63 

64 

65 

66 
67 

68 
69 
70 
71 

72 
73 
74 
75 
76 
77 
78 
79 
80 
81 
82 



37 
20 



33 

20 

3 

7 

47 
56 



7 
8 

19 

20 



2 
5 

145 

29 
15 



34 



23 



2 
3 

5 

24 



39 



1 

To 



62 
21 



439 



0.3 



1) 24 

2) -1 

3) 5 

4) 180 

5) 4 

6) 8 
7)1 

8) 8 

9) 6 



20) 

21) -18 



Answers - Order of Operation 

10) -6 19) 3 

11) -10 

12) -9 

13) 20 

14) -22 22) -3 

15) 2 23) -4 

16) 28 

17) -40 

18) -15 25) 2 



24) 3 



0.4 

1)7 
2) 29 
3)1 

4) 3 

5) 23 

6) 14 

7) 25 

8) 46 

9) 7 

10) 8 

11) 5 

12) 10 

13) 1 

14) 6 

15) 1 

16) 2 

17) 36 

18) 54 



Answers - Properties of Algebra 



19; 


7 


20; 


38 


21; 


r + 1 


22; 


-Ax -2 


23; 


2n 


24; 


116+7 


25; 


\hv 


26; 


7x 


27; 


-9x 


28; 


-7a -1 


29; 


k + 5 


30; 


-3p 


31; 


-5a; -9 


32; 


-9-10n 


33; 


— m 


34; 


— 5 — r 


35; 


10n + 3 


36; 


56 



37; 


- 8x + 32 


38; 


24t> + 27 


39; 


8n 2 + 72n 


40; 


5 -9a 


41; 


-7k 2 + 42k 


42; 


10a; + 20a; 2 


43; 


— 6 — 36a; 


44; 


-2n-2 


45; 


40m — 8m 2 


46; 


- 18p 2 + 2p 


47; 


- 36x + 9a; 2 


48; 


32n-8 


49; 


-9fe 2 + 90fe 


50; 


- 4 - 28r 


51; 


- 40n - 80n 2 


52; 


16x 2 -20x 


53; 


146 + 90 



440 



54) 60t; - 7 


64) 30r - 16r 2 


74) 2x 2 - 6a; - 3 


55) -3x + 8x 2 


65) -72n-48-8n 2 


75) Ap - 5 


56) -89x + 81 


66) -A2b-A5-Ab 2 


76) 3x 2 + 7a; - 7 


57) -68k 2 -8k 


67) 79 - 79t> 


77) -t> 2 + 2v + 2 


58) -19 -90a 

59) -34-49p 


68) -8x + 22 

69) -20n 2 + 80n-42 


78) -7b 2 + Zb-l 


60) -10a; + 17 


70) -12 + 57o + 54o 2 


79) -Ak 2 + 12 


61) 10 -An 


71) -75-20A; 


80) a 2 + 3a 


62) -30 + 9m 


72) -128a; -121 


81) 3x 2 -15 


63) 12x + 60 


73) 4n 2 - 3n - 5 


82) -n 2 + 6 



1.1 

1)7 

2) 11 

3) -5 

4) 4 

5) 10 

6) 6 

7) -19 

8) -6 

9) 18 

10) 6 

11) -20 

12) -7 

13) -108 

14) 5 

1.2 



Answers - Chapter 1 



Answers to One-Step Equations 



15) -8 


29) 5 


16) 4 


30) 2 


17) 17 

18) 4 

19) 20 


31) -11 

32) -14 


20) -208 


33) 14 


21) 3 


34) 1 


22) 16 


35) -11 


23) -13 

24) -9 

25) 15 

26) 8 


36) -15 

37) -240 

38) -135 


27) -10 


39) -16 


28) -204 


40) -380 


swers to Two 


-Step Equations 


2) 7 





441 



3) -14 

4) -2 

5) 10 

6) -12 

7) 

8) 12 

9) -10 

10) -16 

11) 14 

12) -7 

13) 4 

14) -5 

15) 16 



16) 


-15 


17) 


7 


18) 


12 


19) 


9 


20) 





21) 


11 


22) 


-6 


23) 


-10 


24) 


13 


25) 


1 


26) 


4 


27) 


-9 


28) 


15 



29) -6 

30) 6 

31) -16 

32) -4 

33) 8 

34) -13 

35) -2 

36) 10 

37) -12 

38) 

39) 12 

40) -9 



1.3 



Answers to General Linear Equations 



1) -3 


16) 


2) 6 


17) 2 


3) 7 


18) -3 


4) 


19) -3 


5)1 


20) 3 


6) 3 


21) 3 


7) 5 


22) -1 


8) -4 


23) -1 


9) 


24) -1 


10) 3 


25) 8 


11)1 


26) 


12) All real numbers 


27) -1 


13) 8 


28) 5 


14) 1 


29) -1 


15) -7 


30) 1 



31) 




4 


32) 







33) 




3 


34) 







35) 







36) 




2 


37) 




6 


38) 




3 


39) 


5 




40) 


6 




41) 







42) 




2 


43) 


No Solution 


44) 








442 



45) 12 


48) 1 




46) All real numbers 


49) -9 




47) No Solution 


50) 




1.4 








Answers to Solving 


with Fractions 


ol 


11) 


22) -1 


2)-! 


12)| 


23) -2 


3)f 


13) -| 


24) -J 


<>i 


")s 


25) 16 


5)"? 


15) -i 


26) -i 


„, 


16) 1 




6 >- 


17) 


27) -f 


*>-; 


18) -1 


28) -1 


8)-! 


19) 1 




9) -2 


20) 1 


29)1 


io)| 


21) J 


30)| 



1.5 



1. 


b= c - 

a 


2. 


h = gi 


3. 


gb 
X- / 


4. 


»=? 


5. 


a 


6. 


_ cb 
" dm 


7. 




8. 


D = f 


9. 


_ 3V " 
4 r 3 


10 


2£ 

i. m = — 

U2 



Answers - Formulas 
11. c = b-a 22. x= c ~ fe 

12. X = B + f 



cm + en 



17. n : 



p — c 



23. 771 = 



P-3 

13. y = ^^ *»• " b - — 

14. r = M^l 24. L = ^±^ 

& 6 

K n _ 12V 

i5 - ^-^T 25. k = qr 

16. k 



in 



R — L r\fy rp R — b 



18. L = S-2B 27. v = — — 



28. /l = 



19. D = TL + d 

20. £ a = IR + Eg 

21. L = -A- 29. Q 2 = ^^ 



443 



30. n = 


L-2d- nr 2 


TV 


31. 7\ 


Rd - kAT 2 
kA 


32. t) 2 : 


_ Pg + V? 
Vi 


33. a = 


_ c-b 

X 


34. r = 


_ d 
' 1 


35. u; = 


V 
~ Ih 


36. /i = 


3v 

nr 2 


1.6 




1)8, - 


-8 


2)7,- 


7 


3) 1,- 


1 


4) 2, - 


-2 


5) 6, - 


29 


6)f." 


-6 


7) -2 


10 
' ~3~ 


8) -3 


,9 


9) 3, - 


39 

~ T 


10) f, 


-6 


11) 7, 


29 
~~ ~3~ 


12) - 


i-» 


13) - 


9,15 


1.7 





2) 


X 


--k 


3) 


WX: 


= k 


4) 


r 

1 2 = 


--k 


5) 


xy 


--k 



O 1 . 


a — — i — 




38. 


b= c ~ 1 

a 


39. 


, 5 + bw 
a 


40. 


at — s 


W b 


41. 


c— bx 


X 


42. 


x = 3 — by 


43. 


3 — x 

y= 5 


44 


_ T _7-2, 



c — 1 , r 7 — 3x 

45. 2/ = -2- 

46. a = — — 

5 



5a -4 



47. b = ^ 



48. a: = 



8 + 5y 
4 

4x-8 



49. y = - 5 

50. /= 9G + 160 



o 



Answers to Absolute Value Equations 

14) 3,- | 27) 6, -f 

15 >- 2 '° 28)|,0 



16) 0,-2 

17) -fo 29) -^,1 



3 

17 7 



22) 6. 2 ' 



13 



18) -4,i 30) -3,5 



!9) "T^ 31) -i,-f 

20) -|, -2 x 2 

' 5 ' 32) -6, | 

21) -6,-8 



33) 7 i 



3 

23)1,-H 34)-f,-A 

24) 7,-21 

351 -- -- 

25) -2,10 ; 22 ' 38 

26) -J,l 36)0,-f 



Answers - Variation 
6) jm 3 = A; n) Z = o.5 

7 rq 

fc 



12) cd = 28 



a\/b 

9) ab = fc 

10) f = 3 14) 



13) 4 = 0.67 



444 



15) wx 3 = 


1458 


23) 


241,920,000 cans 


32) 


1600 km 




16) - = 1.5 

17) " -0.33 


24) 
25) 
26) 


3.5 hours 
4.29 dollars 
450 m 


33) 
34) 


r = 36 

8.2 mph 




18) mn = 


3.78 


27) 


40 kg 


35) 


2.5 m 




19) 6 k 

20) 5.3 k 

21) 33.3 cm 


28) 
29) 
30) 


5.7 hr 
40 lb 
100 N 


36) 

37) 


V = 100.5 
6.25 km 


cm 3 


22) 160 kg/cm 3 


31) 


27 min 


38) 


I = 0.25 




1.8 


















Answer Set - Number and Geometry 






1)11 




17) 


64, 64, 52 


33) 


40, 200 




2) 5 




18) 


36, 36, 108 


34) 


60, 180 




3) -4 

4) 32 




19) 
20) 


30, 120, 30 
30, 90, 60 


35) 


20, 200 




5) -13 




21) 


40, 80, 60 


36) 


30, 15 




6) 62 




22) 


28, 84, 68 


37) 


76, 532 




7) 16 




23) 


24, 120, 36 


38) 


110, 880 




8)? 

9) 35, 36, 


37 


24) 
25) 


32, 96, 52 
25, 100, 55 


39) 


2500, 5000 




10) -43, 


-42,-41 


26) 


45, 30 


40) 


4, 8 




11) -14, 


-13,-12 


27) 


96, 56 


41) 


2, 4 




12) 52, 54 




28) 


27, 49 


42) 


3, 5 




13) 61, 63 

14) 83, 85 


, 65 

, 87 


29) 
30) 


57, 83 
17, 31 


43) 


14, 16 




15) 9, 11, 


13 


31) 


6000, 24000 


44) 


1644 




16) 56, 56 


, 68 


32) 


1000, 4000 


45) 


325, 950 




1.9 




Ar 


iswers - Age Problems 








1) 6, 16 




4) 


17, 40 








2) 10, 40 




5) 


27, 31 








3) 18, 38 




6) 


12, 48 









445 



7) 31, 36 

8) 16, 32 

9) 12, 20 

10) 40, 16 

11) 10, 6 

12) 12, 8 

13) 26 

14) 8 

15) 4 

16) 3 

17) 10, 20 

18) 14 
1.10 

1)1 4 



2) 25|,20| 


3) 3 


4) 10 


5) 30, 45 


6) 3 


7 n 300 
> 13 


8) 10 


9) 7 


10) 30 


11) 150 


12) 360 


13) 8 



19) 


9, 18 


20) 


15, 20 


21) 


50, 22 


22) 


12 


23) 


72, 16 


24) 


6 


25) 


37, 46 


26) 


15 


27) 


45 


28) 


14, 54 


29) 


8, 4 


30) 


16, 32 



31) 10, 28 

32) 12,20 

33) 141, 67 

34) 16, 40 

35) 84, 52 

36) 14, 42 

37) 10 

38) 10, 6 

39) 38, 42 

40) 5 



Answers - Distance, Rate, and Time Problems 



14) 


10 


15) 


2 


16) 


3 


17) 


48 


18) 


600 


19) 


6 


20) 


120 


21) 


36 


22) 


2 


23) 


570 


24) 


24, 18 


25) 


300 


26) 


8, 16 


27) 


56 



28) 95, 120 

29) 180 

30) 105, 130 

31) 2:15 PM 

32) 200 

33) i 

34) 15 

35)? 

m\ 

37) 3, 2 

38) 90 



2.1 



Answers - Chapter 2 



446 



Answers - Points and Lines 



1) B(4, -3) C(l, 2) D(-l,4) 
E(-5,0) F(2, -3) G(l, 3) 
H(-l,-4) I(-2,-l)J(0, 2) 
K(-4,3) 



2) 



H. 



J. 



D 



K. 



E 



G 



3) 













































































































































4) 













































































































































6) 
























































































































P 





7) 


















































\ 




























































\ 











9) 













































































































































10) 















































































































































13) 













































































































































14) 



447 



s 



16) 

















































































































































=t= 


====== ===$ 


==\=: 






22) 
















==========£ 




:=E 









20) 







































































































































2.2 



1) 3 

' 2 



2) 5 

3) Undefined 



4) - 


i 

~ 2 


5 )| 




6) - 


2 
~ 3 


7) - 


-1 


8)f 




9) - 


-1 


10) 





11) 


Undefined 


12) 


16 

T 


13) 


17 
~~ 31 


14) 


3 

~ 2 



15 

16 
17 
18 
19 
20 
21 

22 
23 
24 
25 
26 
27 



Answers - Slope 



7 
17 



5 
IT 

1 

2 

1 
16 

11 

2 

12 
_ 31 

Undefined 

2-1 

IT 
_ 26 

_ 27 
_ 19 

— To 

1 

_ 3 



28 
29 
30 
31 
32 
33 
34 
35 
36 
37 
38 
39 
40 



i 

16 



7 
IT 



2.3 



448 



Answers - Slope-Intercept 



i) y = 


2x + 5 


2) y = 


-6a; + 4 


3) y = 


x-4 


4) y = 


-x-2 


5) y = 


3 i 
--x-1 


6) y = 


-^x + 3 


7) y = 


ix+1 


8) */ = 


2 , E 

-x + 5 

5 


9) y = 


— x + 5 


10) y = 


7 r 

= - -x - 5 


11) y: 


= x — 1 


12) 2/: 


5 o 

= --x-3 


13) y = 


= -4x 


14) y- 


= -\x + 2 


15) y = 


1 37 

~ ~~ To x ~~ lo 


16) y = 


1 3 

~To x — To 


17) y- 


= - 2x - 1 


18) 2/ = 


6 70 

~ 11 X+ 11 


19) y- 


= 7 -x-8 


20) 2/ = 


= - yX + 4 


21) z = 


= -8 


22) y- 


= y£ + 6 


23) j/ = 


= — x — 1 


24) y = 


5 

= 2 X 


25) t/ = 


= 4x 


26) y = 


= -fx + l 


27) y- 


= - 4x + 3 


28) a; = 


= 4 


2.4 





29) y = 
30)2/ : 
31) 



-x- 



1 



-x + 4 




38) 













































































































































39) 



33) 













































































































































40) 




35) 















































































































































449 



1) x = 2 

2) x=l 

3) y-2= 1 -(x-2) 

4) y-l = - 1 -( x -2) 

5) y + 5 = 9( x+ i) 

6) y + 2 = -2(x-2) 

7) 2/-l = |(x + 4) 

8) y + 3 = -2(x-4) 

9) y + 2 = -3x 

10) y-l = 4(x+l) 



11) y + 5-. 
12)y-2-. 



-x 



-x 



13) y + Z = \{x + h) 

14)y + 4 = - 2 -(x + l) 
15)y-4 = - 5 -(x + l) 

16) y + 4 = -|x(x-l) 

17) y = 2x-3 

18) y = -2x + 2 



.ns 1 


ivers - Point- Slope Form 






19; 


y = - \x + 2 


37; 


y + 2 = |(x + 4) 


20; 


2 10 

y=~3 x ~T 


38; 


j/-l = |(x + 4) 


2i; 


y = \x + 3 


39; 


2/ _5 = I( x _3) 


22; 


y = - \x + 4 


40; 


y + 4 = -(x + l) 


23; 


3 , /i 

y = - -x + 4 


41; 


t/ + 3 = -y(x-3) 


24; 
25; 

26; 

27; 


5 K 

y = - -x - 5 

2 K 

t/ = - -x - 5 

2/ = Jz - 4 

2/ = x — 4 


42; 
43; 
44; 


t/ + 5 = -|(x + l) 

3 11 
y 4 4 

1 3 

^ ~ w x 2 


28; 


y = -3 


45; 


8 5 

V = — -=x — - 


29; 
30; 


x = — 3 
2/ = 2x — 1 


46; 


1 3 

y=2 x ~2 


31; 


y = -^ 


47; 


y = — x + 5 


32; 


6 
y = -x - 3 


48; 


y = \x+l 


33; 


y-3 = -2(x + 4) 


49; 


y = — x + 2 


34; 


Z/ = 3 


50; 


y = x + 2 


35; 


y-l = ^(x-5) 


si; 


y = 4x + 3 


36; 


y-5 = -|(x + 4) 


52; 


3 . 6 

y = 7 X + 7 



2.5 

1)2 

2) - 

3) 4 

4) - 
5)1 



10 
T 



6)1 

7) -7 

8) -! 



Answers - Parallel and Perpendicular Lines 

9)0 17) x = 2 

10)2 18)y-2= 7 -(x-5) 

11) 3 

12) ~\ 

13) -3 

">-! 

15) 2 



16) 



19) 2/-4 = |(ar-3) 

20) 2/ + l = -|(x-l) 

21) 2/-3 = J(x-2) 

22) y-3 = -3(x + l) 



450 



23) x = 4 




32)y-5 = - 1 -(x 


+ 2) 


41) 


y = x-l 


2A)y-A = \{x-l) 




33) y = -2x + 5 




42) 


y = 2x+l 


25) y + 5 = -(x-l) 

26) y + 2 = -2(x-l) 

27)y-2= 1 -(x-5) 
28) y-3 = -(x-l) 




34) y = lx + 5 

35) t/ = — -x — 3 

36) y = - \x - 5 

37) y = - \x - 3 




43) 
44) 
45) 


y = 2 

y = - -x + 1 

y = — x + 3 


29)y-2 = - 1 -(x-4) 




38)y = fx-2 




46) 


y = - \x + 2 


30) y + 5 = l(x + S) 




3V)y=-\x-2 




47) 


y = — 2x + 5 


31) y + 2 = -3(x-2) 




40) y = \x-\ 




48) 


3 l A 




A 


nswers - Ch 


apter 


3 





3.1 



Answers - Solve and Graph Inequalities 



1) (-5,oo) 

2) (-4,oo) 

3) (-oo,-2] 

4) (-oo,l] 

5) ( — oo,5] 

6) ( — 5, 00) 

7) m < - 2 

8) ra<l 

9) x ^5 

10) a^- 5 

11) 6>-2 

12) col 

13) x ^ 110: [110, 00) 

14) n^ -26: [-26, 00) 

15) r<l:(-oo,l) 

16) m ^ — 6: ( — 00, — 6] 

17) ra>-6:[-6,oo) 



18 


) x < 6: ( — 00, 6) 


19 


) a<12:(-oo,12) 


20 


) v ^ 1: [1, 00) 


21 


) x^ll: [ll,oo) 


22 


) x<-18:(-oo,-18] 


23 


) fc>19:(19,oo) 


24 


) n^- 10: (-00,- 10] 


25 


) p<- 1: (-00,- 1) 


26 


) x^ 20: (-00, 20] 


27 


) m^2: [2,oo) 


28 


) n ^ 5: ( — 00, 5] 


29 


) r>8:(8,oo) 


30 


) x ^ — 3: ( — 00, — 3] 


31 


) 6>l:(l,oo) 


32 


) n^0:[0,oo) 


33 


) «<0:(-oo,0) 


34 


) x>2:(2,oo) 



451 



35) No solution: 37) {All real numbers.} : R 

36) n>l:(l,oo) 38) p< 3: ( - oo, 3] 

3.2 

Answers - Compound Inequalities 

1) n^-9orn^2:(-oo,-9]U [2, oo) 

2) m^-4orm< -5 : (- oo,-5)|J [-4, oo) 

3) x^5oi x < — 5: ( — oo, — 5) |J [5, oo) 

4) r>0orr<-7 : (-oo-7),|J (0, oo) 

5) x<-7: (-oo,-7) 

6) n<-7orn>8: (-00-7), (J (8, oo) 

7) -8<t><3:(-8,3) 

8) -7<x<4:(-7,4) 

9) 6<5:(-oo,5) 

10) -2<nsC6:[-2,6] 

11) -7<a<6:[-7,6] 

12) O6:[6,oo) 

13) -6^x^-2: [-6, -2] 

14) -9<x<0:[-9,0] 

15) 3<fcsC4:(3,4] 

16) -2<nsC4:[-2,4] 

17) -2<x<2:(-2,2) 

18) No solution :0 

19) -l<m<4:[-l,4) 

20) r>8orr<-6:: ( - oo, - 6) (J (8, oo) 

21) No solution :0 

22) rcsC0or:c>8: (-oo,0]|J (8, oo) 

23) No solution : 

24) n ^ 5 or n < 1 : ( - oo, 1) |J [5, oo) 



452 



25) 5<x<19:[5,19) 

26) n<-Uorn^l7: ( - oo, - 14) (J [17, oo) 

27) l^v<8:[l,8] 

28) a<lora^l9: ( - oo, 1] (J [19, oo) 

29) k ^ 2 or k < - 20 : ( - oo, - 20) U [2, oo) 

30) {All real numbers.} : R 

31) -l<x^l:(-l,l] 

32) m > 4 or m ^ - 1 : ( - oo, - 1] \J (4, oo) 



3.3 

1) - 

2) - 

3) ■ 

4) - 

5) - 

6) - 

7) 
8) 

9) 
10) 

11) 
12) 
13) 
14) 

15) 

16) 
17) 



-3,3 

-8,8 

-3,3 

-7,1 

-4,8 

-4,20 

-2,4 

-7,1 

_ 7 11 
_ 3 ' T 

-7,2 

-3,5 

0,4 

1,4 

(-oo,5) (J (5,oo 

( — oo 

( — oo 
( — oo 



Answers - Absolute Value Inequalities 

18) (-oo,-l)|J (5,oo) 



|]U [f,o°) 

1](J [9,oo) 
6)U (0,oo) 



19)(-oo,|)U(|,oo) 

20) (-oo,0)U (4,oo) 

21) (-oo,-l]|J [3,oo) 

22) [-|,2] 

23) (-oo, -4] (J [14, oo) 

24)(-oo,-|U [-|,oo) 

25) [1,3] 

26) [i 1] 

27) (-oo,-4)|J (-3,oo) 

28) [3,7] 

29) [1,|] 

30) [-2,-|] 
31)(-oo,|)U(|,oo) 

32) (-oo,-|)U (l,oo) 

33) [2,4] 

34) [-3,-2] 



453 



4.1 



Answers - Chapter 4 



Answers - Graphing 



1) 


(-1, 


2) 


2) 


(-4, 


3) 


3) 


(-1, 


-3) 


4) 


(-3, 


1) 


5) 


Mo Solution 


6) 


(-2, 


-2) 


7) 


(-3, 


1) 


8) 


(4,4) 




9) 


(-3, 


-1) 


10) 


No Solution 


11) (3,- 


-4) 



4.2 



1) (1,-3) 


2) (-3,2) 


3) (-2,-5) 


4) (0,3) 


5) (-1,-2) 


6) (-7,-8) 


7) (1,5) 


8) (-4,-1) 


9) (3,3) 


10) (4, 4) 


11) (2,6) 


12) (-3,3) 


13) (-2,-6) 


14) (0,2) 


4.3 



12) (4,-4) 


23) 


(-1,-1) 


13) (1,-3) 

14) (-1,3) 


24) 


(2,3) 


15) (3,-4) 


25) 


(-1,-2) 


16) No Solution 


26) 


(-4,-3) 


17) (2,-2) 

18) (4,1) 


27) 


No Solution 


19) (-3,4) 


28) 


(-3,1) 


20) (2,-1) 

21) (3,2) 


29) 


(4,-2) 


22) (-4,-4) 


30) 


(1,4) 


Answers - Substitution 






15) (1,-5) 


29) 


(4,-3) 


16) (-1,0) 


30) 


(-1,5) 


17) (-1,8) 


31) 


(0,2) 


18) (3,7) 

19) (2,3) 


32) 


(0,-7) 


20) (8,-8) 


33) 


(0,3) 


21) (1,7) 


34) 


(1,-4) 


22) (1,7) 


35) 


(4,-2) 


23) (-3,-2) 


36) 


(8,-3) 


24) (1,-3) 

25) (1,3) 

26) (2,1) 


37) 
38) 


(2,0) 

(2,5) 


27) (-2,8) 


39) 


(-4,8) 


28) (-4,3) 


40) 


(2,3) 



454 



l)(-2,4) 

2) (2,4) 

3) No solution 

4) Infinite number of 

solutions 

5) No solution 

6) Infinite number of 

solutions 

7) No solution 

8) (2,-2) 

9) (-3,-5) 

10) (-3,6) 

11) (-2,-9) 

4.4 



1) (1,-1,2) 

2) (5,-3,2) 

3) (2,3,-2) 

4) (3,-2,1) 

5) (-2,-1,4) 

6) (-3,2,1) 

7) (1,2,3) 

8) oc solutions 

9) (0,0,0) 

10) oc solutions 

11) (19,0,-13) 
4.5 

1) 33Q, 70D 



5wer 


s - Addition, 


/Elimination 




12) 


(1,-2) 


25) 


:-i,-2) 


13) 


(0,4) 


26) 


:-3,o) 


14) 
15) 


(-1,0) 

(8,2) 


27) 


:-i,-3) 


16) 


(0,3) 


28) 


;-3,o) 


17) 


(4,6) 


29) 


:-s,9) 


18) 


(-6,-8) 


30) 


;i,2) 


19) 
20) 


(-2,3) 
(1,2) 


31) 


:-2,i) 


21) 


(0,-4) 


32) 


:-i,i) 


22) 


(0,1) 


33) 


;o,o) 


23) 


(-2,0) 


34) ] 


mfinite number of 


24) 


(2,-2) 




solutions 



Answers - Three Variables 

12) oc solutions 

13) (0,0,0) 

14) oc solutions 
15)(2,i-2) 

16) oc solutions 

17)(-1,2,3) 

18)(-l,2,-2) 

19) (0,2,1) 

20) no solution 

21) (10,2,3) 

22) no solution 

Answers - Value Problems 



23 

24 
25 
26 
27 

28 
29 
30 
31 
32 



(2,3,1) 

oc solutions 

no solutions 

1,2,4) 

-25,18,-25) 

2 3 _ 2n 
7 ' 7 ' ~~ 7> 

1,-3,-2,-1) 
7,4,5,6) 
1,-2,4,-1) 
-3,-1,0,4) 



455 



2) 26 h, 8 n 

3) 236 adult, 342 child 

4) 9d, 12q 

5) 9, 18 

6) 7q, 4h 

7) 9, 18 

8) 25, 20 

9) 203 adults, 226 child 

10) 130 adults, 70 

students 

11) 128 card, 75 no card 

12) 73 hotdogs, 
58 hamburgers 

13) 135 students, 

97 non-students 

14) 12d, 15q 

15) 13n, 5d 

16) 8 20c, 32 25c 

17) 6 15c, 9 25c 

18) 5 
4.6 



1) 2666.7 

2) 2 

3) 30 

4) 1,8 

5) 8 

6) 10 

7) 20 

8) 16 

9) 17.25 



19) 13 d, 19 q 

20) 28 q 

21) 15 n, 20 d 

22) 20 $1, 6 $5 

23) 8 $20, 4 $10 

2!) 27 

25) $12500 @ 12% 
$14500 @ 13% 

2(i) $20000 @ 5% 
$30000 @ 7.5% 

27) $2500 @ 10% 
$6500 @ 12% 

28) $12400 @ 6% 
$5600 @ 9% 

29) $4100 @ 9.5% 
$5900 @ 11% 

30) $7000 @ 4.5% 
$9000 @ 6.5% 

31) $1600 @ 4%; 
$2400 @ 8% 

32) $3000 @ 4.6% 
$4500 @ 6.6% 



Answers - Mixture Problems 

10) 1.5 

11) 10 

12) 8 

13) 9.6 

14) 36 

15) 40,60 

16) 30,70 

17) 40,20 

18) 40,110 



33) 


$3500 @ 6%; 




$5000 @ 3.5% 


34) 


$7000 @ 9% 




$5000 @ 7.5% 


35) 


$6500 @ 8%; 




$8500 @ 11% 


36) 


$12000 @ 7.25% 




$5500 @ 6.5% 


37) 


$3000 @ 4.25%; 




$3000 @ 5.75% 


38) 


$10000 @ 5.5% 




$4000 @ 9% 


39) 


$7500 @ 6.8%; 




$3500 @ 8.2% 


40) 


$3000 @ 11%; 




$24000 @ 7% 


41) 


$5000 @ 12% 




$11000 @ 8% 


42) 


26n, 13d, lOq 


43) 


18, 4, 8 


44) 


20n, 15d, lOq 



19) 


20,30 


20) 


100,200 


21) 


40,20 


22) 


10,5 


23) 


250,250 


24) 


21,49 


25) 


20,40 


26) 


2,3 



456 



27) 56,144 

28) 1.5,3.5 

29) 30 

30) 10 

31) 75,25 

32) 55,20 



33) 440, 160 

34) 20 

35) 35,63 

36) 3,2 

37) 1.2 

38) 150 

Answers - Chapter 5 



39) 10 

40) 30,20 

41) 75 

42) 20,60 

43) 25 



5.1 



Answers to Exponent Properties 



1)4 9 


17 


)4 2 


31 


) 64 


2) 4 7 


18 


)3 4 


32 


) 2a 


3) 2 4 


19 


) 3 


33 


> y 3 


4) 3 6 


20 


)3 3 




' 512x 24 


5) 12m 2 n 


21 


m 


34 


y 5 s 2 
' 2 


6) 12a; 3 


22 


xy 3 


35 


64m 12 n 


7) 8m 6 n 3 




4 




„10 


8) x 3 y 6 


23 


. 4x 2 j/ 
' 3 


36 


. n 

' 2m 


9)3 12 


24 


) ^ 
' 4 


37 


) 2x 2 y 


10) 4 12 


25 


) 4x w y 14 


38 


)2y 2 


11) 4 8 


26 


) 8u 18 v e 


39 


2q 7 r 8 p 


12) 3 6 


27 


) 2x 17 y 16 


40 


) 4x 2 t/S 2 


13) 4uV 










/ 


28 


) 3uv 


41 


4 16 4 

) x y z 


14) x 3 y 3 




2 






15) 16a 16 


29 


x y 
' ~6~ 


42 


) 256g 4 r 8 


16) 16xV 


30 


4a 2 
' ~3~ 


43 


) 4y 4 z 


5.2 












Answc 


;rs to Negative 


Exponents 




1) 32x 8 y w 


3) 


2a 15 






2) 32 f 


4) 


2x 3 y 2 







457 



5) 16:r 4 y s 

6) 1 

7) y 16 x 5 



32 



9) 
10) 



li: 



!);,/ 



2x 7 



y .1, 

12) ^f 



14) 



15) 



*V 



12-u 5 



17 >? 



5.3 



1) 8.85 x 10 2 

2) 7.44 x 10" 4 

3) 8.1 xlO" 2 

4) 1.09 x 10° 

5) 3.9 xlO" 2 

6) 1.5 x 10 4 

7) 870000 

8) 256 

9) 0.0009 

10) 50000 

11) 2 

12) 0.00006 

13) 1.4 x 10" 3 

14) 1.76 x 10" 10 



18) 


a 
~2b~ 


19) 


16a 12 b 12 


20) 


y s x 4 
4 


21) 


1 

8m 4 n 7 


22) 


2x 16 y 2 


23) 


16n 6 m A 


24) 


2x 


25) 


1 

x 15 y 


26) 


4y 4 


27) 


u 

2^ 


28) 


4 y 5 


29) 


8 


30) 


i 

2l/.3„5 



31) 


2y 5 x* 


32) 


a 3 
263 


33) 


1 


x 2 y 11 z 


34) 


a 2 


8c 10 6 12 


35) 


1 

h 3 kj 6 


36) 


x 30 z 6 
16y 4 


37) 


26 14 
a 12 c 7 


38) 


14 8 

m q 


4p 4 


39) 


X 2 

y 4 z 4 


40) 


mn 7 



Answers to Scientific Notation 



15 


) 1.662 


x 10" 6 


16 


) 5.018 


x 10 6 


17 


) 1.56 


x 10- 3 


18 


) 4.353 


x 10 8 


19 


) 1.815 


xlO 4 


20 


) 9.836 


x 10" 1 


21 


) 5.541 


x 10" 5 


22 


) 6.375 


xlO" 4 


23 


) 3.025 


x 10" 9 


24 


) 1.177 


x 10" 16 


25 


) 2.887 


x 10" 6 


26 


) 6.351 


x lO" 21 


27 


) 2.405 


x lO" 20 


28 


) 2.91 x 


lO" 2 



29 
30 
31 
32 
33 
34 
35 
36 
37 
38 
39 
40 
41 



1.196 x 10" 2 
1.2 x 10 7 
2.196 x 10" 2 
2.52 x 10 3 
1.715 x 10 14 
8.404 x 10 1 
1.149 x 10 6 
3.939 x 10 9 
4.6 x 10 2 
7.474 x 10 3 
3.692 x 10" 7 
1.372 x 10 3 
1.034 x 10 6 



458 



42) 1.2x10 



(> 



5.4 



Answers to Introduction to Polynomials 



1)3 


16) 


2w 4 + 6 






31) n 3 -5n 2 + 3 


2) 7 


17) 


13p 3 






32) -6a; 4 +13x 3 


3) -10 


18) 


— 3a; 






33) -12n 4 + n 2 + 7 


4) -6 


19) 


3n 3 + 8 






34) 9a; 2 + 10a; 2 


5) -7 


20) 


x 4 + 9a; 2 - 


-5 






6) 8 


21) 


26 4 + 26+10 




35) r 4 - 3r 3 + 7r 2 + 1 


7)5 


22) 


-3r 4 +12r 2 -l 




36) 10x 3 - 6a; 2 + 3x - 8 


8) -1 


23) 


- 5a; 4 + 14a; 3 - 1 




37) 9n 4 + 2n 3 + 6n 2 


9) 12 


24) 


5n 4 — An + 7 
















38) 26 4 - b 3 + 46 2 + 46 


10) -1 


25) 


7a 4 - 3a 2 


-2a 






11) 3p 4 -3p 


26) 


12w 3 + 3w + 3 




39) - 36 4 + 136 3 - 76 2 












116+19 


12) -m 3 +12m 2 


27) 


p 2 + 4p — 


6 




40) 12n 4 - n 3 - 6n 2 + 10 


13) -n 3 +10n 2 


28) 


3m — 2m 


+ 6 






14) 8a; 3 + 8x 2 


29) 


56 3 +126 2 


+ 5 




41) 2x 4 -x 3 -4x + 2 


15) 5n 4 + 5n 


30) 


- 15n 4 + 4n - 6 




42) 3a; 4 + 9a; 2 + 4a; 


5.5 














Answers to Multiply Polynomials 


1) 6p-42 






13; 


15v 2 


-26f + 8 


2) 32k 2 + 16k 






14; 


6a 2 - 


- 44a - 32 


3) 12a; + 6 






!s; 


24a; 2 


- 22x - 7 


4) 18n 3 + 21n 2 


















i6; 


20a; 2 


-29x + 6 


5) 20m 5 + 20m 4 


















!7; 


30a; 2 


— lAxy — Ay 2 


6) 12r-21 


















is; 


16m 2 


+ \Quv-2lv 2 


7) 32n 2 + 80n + 48 












8) 2x 2 - 7a; - 4 






19; 


3x 2 + 13a;y + 12 if 


















20; 


40m 2 


-34uv-48v 2 


9) 56& 2 — 196 — 15 












10) 4r 2 + 40r+64 






21; 


56a- 2 


+ 61xy + 15y 2 


11) 8a; 2 + 22x + 15 






22; 


5a 2 - 


-7a6-24fo 2 


12) 7n 2 + 43n-42 






23; 


6r 3 - 


-43r 2 + 12r-35 



459 



24) 16x 3 + 44x 2 + 44x + 40 

25) 12n 3 -20n 2 + 38n-20 

26) 8b 3 - 4b 2 - 4b - 12 

27) 36a; 3 - 24x 2 y + Sxy 2 + 12y 3 

28) 21m 3 + 4m 2 n-8n 3 

29) 48n 4 - 16n 3 + 64n 2 - 6n + 36 

30) 14a 4 + 30a 3 - 13a 2 - 12a + 3 

31) 15fc 4 + 24k 3 + 48A; 2 + 27k + 18 



32) 42u 4 + 76u 3 v + 17u 2 v 2 - 18i> 4 

33) 18x 2 -15x-12 

34) 10:r 2 -55:c + 60 

35) 24x 2 - 18x - 15 

36) 16x 2 -44x-12 

37) 7x 2 - 49x + 70 

38) 40x 2 -10x-5 

39) 96x 2 - 6 

40) 36a; 2 + 108a; + 81 



5.6 



1) a; 2 -64 

2) a 2 -16 

3) 1 - 9p 2 

4) x 2 -9 

5) 1 - 49n 2 

6) 64m 2 - 25 

7) 25n 2 -64 

8) 4r 2 - 9 

9) 16a; 2 -64 

10) b 2 - 49 

11) 16y 2 -a; 2 

12) 49a 2 -49b 2 

13) 16m 2 - 64n 2 

14) 9y 2 -9a; 2 



Answers 


to Multiply Sped 


15) 


36x 2 -4y 2 


16) 


l + 10n + 25n 2 


17) 


a 2 + 10a + 25 


18) 


v 2 + 8w + 16 


19) 


x 2 - 16a; + 64 


20) 


1- 12n + 36n 2 


21) 


p 2 + 14p + 49 


22) 


49fc 2 -98fc + 49 


23) 


49 - 70n + 25n 2 


24) 


16a; 2 - 40a; + 25 


25) 


25m 2 - 80m + 64 


26) 


9a 2 + 18a6 + 96 2 


27) 


25a; 2 + 70a;y + 49y 2 


28) 


16m 2 — 8m n + n 2 



29) 


4a; 2 + 8a;y + 4y 2 


30) 


64ai 2 + 80a;y + 25y 2 


31) 


25 + 20r + 4r 2 


32) 


m 2 - 14m + 49 


33) 


4 + 20a; + 25a; 2 


34) 


64n 2 - 49 


35) 


16w 2 - 49 


36) 


6 2 -16 


37) 


n 2 -25 


38) 


49a; 2 + 98a; + 49 


39) 


16fc 2 +16fc + 4 


40) 


9a 2 - 64 



5.7 



1) 5x 



2) 



5.ir 



5x z 



It 



Answers to Divide Polynomials 
3) 2n 3 

/ 8 ' 9 



"' ' Io +4n 



460 



5) 2x 3 + Ax 7 



np' 



6) ^ + 4p 2 + 4p 

7) n 2 + 5n + \ 



2m + 3 



9) x -10 



i 



10) r + 6 

11) n + 8 

12) 6-3 

13) w + 8 

15)a + 4- 

16) x — 6 - 

17) 5p + 4 

18) 8* -9 ,,_, 



r-9 

8 
n + 5 

5 
b-7 

9 

D-10 

5 



a-8 

2 
x -4 

3 



9p + 4 
1 



19 


i T 1 I d 


' X J ' 10z-2 


20 


) n + 3 + ^- 

n + 4 


21 


U 1 | 2 


' r X ' 4x + 3 


22 


\ rn \ A \ 


m — 1 


23 


) n + 2 


24 


\ T A _i_ 


' X M 2x + 3 


25 


1 Oh 1 5 5 


' J ° ' ° 3fe + 8 


26 


) 7' 1 3 5 


' V ' d 3^-9 


27 


i 7 7 


4x - 5 


28 


, rj 3 


' " ' 4n + 5 


29 


)a 2 +8a 7 a ° ? 


30 


)8£ 2 2k 4 +fc ! 8 


31 


) x 2 4x 10 \ 

x + 4 


32 


) x 2 - 8x + 7 



OOJ 


OIL — i)'IL 


— 1U 


n + 6 


34) 


k 2 -3k- 


-9- 


5 
k-l 


35) 


x 2 - 7x + 3 + 


1 

x + 7 


36) 


n 2 + 9n - 


-1 + 


3 
2n + 3 


37) 


p 2 + 4p - 


-1 + 


4 
9p + 9 


38) 


m 2 — 8m 


+ 7- 


7 


8m + 7 


39) 


r 2 + 3r — 


4 


8 
r — 4 


40) 


x 2 + 3x - 


-7 + 


5 
2a; + 6 


41) 


6n 2 — 3n 


3+ 2n + 3 


42) 


6b 2 + b + 9 + - 


3 
16 — 7 


4^1 


ii — (\ij -+ 


-fi + 


1 



4i> + 3 



Answers - Chapter 6 



6.1 

1) 9 + 86 2 

2) x-5 

3) 5(9x 2 -5) 

4) 1 + 2n 2 

5) 7(8 -op) 

6) 10(5x-8j/) 

7) 7o6(l-5o) 

8) 9x 2 y 2 (Sy 3 - 



ix) 



Answers - Greatest Common Factor 

9) Sa 2 b(-l + 2ab) 

10) 4x 3 (2y 2 + l) 

11) -5rc 2 (l + £ + 3a: 2 ) 

12) 8n 5 (-4n 4 + 4n + 5) 

13) 10(2x 4 -3a; + 3) 

14) 3(7p 6 + 10p 2 + 9) 

15) 4(7m 4 + 10m 3 + 2) 



461 



16) 2x(-5x 3 + 10x + 6) 

17) 5(6fe 9 + ab - 3a 2 ) 

18) 3y 2 (9y 5 + 4x + 3) 

19) -8a 2 b(6b + 7a + 7a 3 ) 

20) 5(6m 6 + 3mn 2 -5) 

21) 5x 3 y 2 z(4x 5 z + 3x 2 + 7y) 

22) 3(p + 4q-5q 2 r 2 ) 

23) 10(5x 2 y + y 2 + 7x2 2 ) 



24) 10y 4 2 3 (3x 5 + 52 2 -x) 

25) 5g(6pr — p+ 1) 

26) 76(4 + 2b + 56 2 + 6 4 ) 

27) 3(-6n 5 + n 3 -7n + l) 

28) 3a 2 (10a 6 + 2a 3 + 9a + 7) 

29) 10x n ( - 4 - 2x + 5x 2 - 5x 3 ) 

30) Ax 2 (-6x 4 -x 2 + 3x + l) 

31) 4mn( - 8n 7 + m 5 + 3n 3 + 4) 

32) 2y\ - 5 + 3t/ 3 - 2xy 3 - 4xy) 



6.2 



1 


) (8r 2 -5)(5r-l) 


2 


) (5x 2 -8)(7x-2) 


3 


) (n 2 -3)(3n-2) 


4 


) (2v 2 -l)(7v + 5) 


5 


) (3b 2 -7)(5b+7) 


6 


) (6a: 2 + 5) (a; -8) 


7 


) (3s 2 + 2)(a; + 5) 


8 


) (7p 2 + 5)(4p + 3) 


9 


) (7s 2 -4)(5x-4) 



10)(7n 2 -5)(n + 3) 



6.3 



1) (p + 9)(p + 8) 

2) (x-8)(x + 9) 

3) (n-8)(n-l) 



Answers - Grouping 



11 
12 
13 
14 
15 
16 
17 
18 
19 
20 



Answers - Trinomials where a = 1 



7x + 5)(y-7) 


21) 


(4u + 3)(8i>-5) 


7r 2 + 3)(6r-7) 


22) 


2(u + 3){2v + 7u) 


8x + 3)(4y + 5x) 
3a + b 2 ){bb-2) 


23) 


(5x + 6)(2y + 5) 


8x + l)(2y-7) 


24) 


(4x-5y 2 )(62/-5) 


m + 5)(3n-8) 


25) 


(3u-7)(v-2u) 


2x + 7y 2 )(y-4x) 
m-5)(5n + 2) 


26) 


(7o-2)(86-7) 


5x-y)(8y + 7) 


27) 


(2x + l)(8t/-3x) 


8x-l){y + 7) 







4) (x-5)(x + 6) 

5) (x + l)(x-10) 

6) (x + 5)(x + 8) 



462 



7) (6 + 8)(o + 4) 


17) 


[u — 5v)(u — 3v) 


8) (6-10)(6-7) 


18) 


{m + 5n)(m — 8n) 


9) (ar-7)(a;+10) 


19) 


[m + 4n)(m — 2n) 


10) (x-3)(x + 6) 


20) 


[x + 8y)(x + 2y) 


11) (n-5)(n-3) 


21) 


[x — 9y)(x — 2y) 


12) (a + 3)(o-9) 


22) 


[u — 7v)(u — 2v) 


13) (p + 6)(p + 9) 


23) 


[x — 3y)(x + 4y) 


14) (p+10)(p-3) 


24) 


[x + 5y)(x + 9y) 


15) (n-8)(n-7) 


25) 


[x + 6y)(x — 2y) 


16) (m — 5n)(m — lOn) 


26) ' 


l(x + 7)(x + 6) 



27) 5(a + 10)(a + 2) 

28) 5(n-8)(n-l) 

29) 6(a-4)(a + 8) 

30) 5(v-l)(v + 5) 

31) 6(x + 2y)(x + y) 

32) 5(m 2 + 6mn-18n 2 ) 

33) 6(x + 9y)(x + 7y) 

34) 6(m — 9n) (m + 3n) 



6.4 



1 


) (7x- 


6)(x - 


-6) 


2 


) (7ra- 


2)(n- 


-6) 


3 


) (76+l)(6+2) 


4 


) (7w + 


4)(t/- 


-4) 


5 


) (5a + 


7)(o- 


-4) 


6 


Prime 




7 


) (2x- 


l)(s- 


-2) 


8 


) (3r + 


2)(r- 


-2) 


9 


) (2x + 


5)(x + 7) 


1( 


3) (7x 


-6)(x 


+ 5) 


1 


L) (26- 


-3)(6 + l) 


1 


2) (5A; 


-6)(/c 


-4) 


1. 


3) (5A; 


+ 3)(& + 2) 


h 


1) (3r- 


f7)(r 


+ 3) 



6.5 



1) (r + 4)(r-4) 

2) (x + 3)(x-3) 



Answers - Trinomials where a ^ 1 



15 


) (3x-5)(x-4) 


29) (Jfc-4)(4fc-l) 


16 


) (3u-2v)(u + 5v) 


30) (r-l)(4r + 7) 


17 


) (3x + 2y)(x + 5y) 


31) (x + 2y)(4x + y) 


18 


) (7x + 5y)(x-y) 


32) 2(2m 2 + 3rrm + 3n 2 ) 


19 
20 
21 
22 


) (5x — 7y)(x + 7y) 
) (hu-4v)(u + 7v) 
) 3(2x+l)(a;-7) 

) 2(5a + 3)(a-6) 


33) (m — 3n) (4m + 3n) 

34) 2{2x 2 -3xy + lby 2 ) 

35) (x + 3y)(4x + y) 


23 


) 3(7fc + 6)(A;-5) 


36) 3(3u + Av)(2u-3v) 


24 


) 3(7n-6)(n + 3) 


37) 2(2x + 7y)(3x + 5y) 


25 


) 2(7x-2)(x-4) 


38) 4(x + 3y)(4x + 3y) 


26 


) (r + l)(4r-3) 


39) 4(x-2y)(6x-y) 


27 


) (x + 4)(6x + 5) 


40) 2(3x + 2y)(2x + 7y) 


28 


) (3p + 7)(2p-l) 





Answers - Factoring Special Products 

3) (v + 5)(v-5) 

4) (x + l)(x-l) 



463 



5) ( p +2)(p-2) 


6) (2v+l)(2v-l) 


7) (3k + 2)(3k-2) 


8) (3o + l)(3o-l) 


9) 


3(ar + 3)(x-3) 


10 


) 5(n + 2)(n-2) 


11 


) 4(2x + 3)(2a;-3) 


12 


) 5(25x 2 + 9y 2 ) 


13 


) 2(3a + 56)(3a-56) 


14 


) 4(m 2 + 16n 2 ) 


15 


) (a-1) 2 


16 


) (k + 2) 2 


17 


) (x + 3) 2 


18 


) (n-4) 2 


19 


) (x-3) 2 


20 


) (k-2) 2 


21 


) (5p-l) 2 


22 


) (x + 1) 2 


23 


) (5a + 36) 2 


24 


) (x + 4y) 2 


25 


) (2a -5b) 2 


26 


) 2(3m-2n) 2 



27 


) 2(2x-3y) 2 


28 


) 5(2x + y) 2 


29 


) (2-m)(4 + 2m + m 2 ) 


30 


) (x + 4)(a; 2 -4x + 16) 


31 


) (x-4)(x 2 + 4x + 16) 


32 


) (x + 2)(x 2 -2x + 4) 


33 


) (6-u)(36 + 6u + u 2 ) 


34 


) (5x-6)(25x 2 + 30a; + 36) 


35 


) (5a - 4) (25a 2 + 20a + 16) 


36 


) (4x-3)(16x 2 + 12a; + 9) 


37 


) (4x + 3y)(16x 2 -12xy + 9y 2 ) 


38 


) 4(2m — 3n)(4m 2 + 6mn + 9n 2 ) 


39 


) 2(3x + 5y)(9x 2 - 15xy + 25y 2 ) 


40 


3(5m + 6n)(25m 2 — 30mn + 36n 2 


41 


) (a 2 + 9)(a + 3)(a-3) 


42 


) (x 2 + 16)(x + 4)(x-4) 


43 


) (4 + z 2 )(2 + z)(2-z) 


44 


) (n 2 + l)(n+l)(n-l) 


45 


) (x 2 + y 2 )(x + y)(x-y) 


46 


) (4a 2 + b 2 ) (2a + 6) (2a - o) 


47 


) (m 2 + 96 2 )(m + 36)(m-36) 


48 


) (9c 2 + 4d 2 )(3c + 2d)(3c-2d) 



6.6 

1) 3(2a + 5y)(4z-3h) 

2) (2x-5)(x-3) 

3) (5u - 4v)(u - v) 

4) 4(2x + 3y) 2 



Answers - Factoring Strategy 

6) 5(4u-x)(v-3u 2 ) 

7) n(5n-3)(n + 2) 

8) x(2x + 3y)(x + y) 

9) 2(3n-2)(9n 2 + 6« + 4) 



5) 2(-x + 4y)(x 2 + 4xy + 16y 2 ) 



10) 2(3-4x)(9 + 12x + 16x 2 ) 



464 



11) n(n-l) 






27) 


(3x - 4) (9x 2 +12:c + 16) 


12) (5ar + 3)(a;-5) 






28) 


(4a + 36) (4a -3b) 


13) (x-3y)(x-y) 






29) 


x(5x + 2) 


14) 5(3u-5v) 2 






30) 


2(x-2)(x-3) 


15) (3x + 5s/)(3x-5y) 






31) 


3fc(fc-5)(fc-4) 


16) (x-3y)(x 2 + 3xy + 9y 2 ) 




32) 


2(Ax + 3y)(Ax-3y) 


17) (m + 2n)(m — 2n) 






33) 


(m — Ax)(n + 3) 


18) 3(2a + n)(2o-3) 






34) 


(2fc + 5)(fc-2) 


19) 4(3o 2 + 2x)(3c-2d) 






35) 


(Ax - y) 2 


20) 3m(m + 2n)(m — An) 






36) 


v(v+l) 


21) 2(4 + 3x)(16-12a; + 9x 2 ) 




37) 


3(3m + An) (3m — An) 


22) (4m + 3n)(16m 2 -12mn + 9n 2 ) 




38) 


x 2 (x + A) 


23) 2x(a; + 5y)(a;-22/) 






39) 


3x(3x — 5y)(x + Ay) 


24) (3a + x 2 )(c + 5d 2 ) 






40) 


3n 2 (3n-l) 


25) n(n + 2)(n + 5) 






41) 


2(m — 2n) (m + 5n) 


26) (4m — n)(16m 2 + 4mr 


i + n 2 ) 




42) 


v 2 (2u-5v)(u-3v) 


G.7 












Answers - 


Solve by 


Factoring 


1)7,-2 


13) 4,0 






25) §,- 5 


2) -4,3 
3)1,-4 


14) 8,0 
15)1,4 






26) -if 


4) -|,7 


16) 4,2 






27) -f,-3 


5) -5,5 


17) f,- 


8 




28) -|,-3 


6)4,-8 

7)2,-7 
8) -5,6 


i8) -; 

19) y, - 


-8 
3 




29) -4,1 

30) 2,-3 


9) -|,-3 

10) -J, 8 

11) ~h 2 


20)^,3 
21) -4 

22)8,- 
23)8,- 


-3 

4 
2 




31) -7,7 

32) -4,-6 

33) -|,-8 


12) -J,2 


24) 4,0 






34) -|,-7 



465 



35) 



5' 



36) §,- 2 

Answers - Chapter 7 



7.1 
1)3 

4) undefined 

6) 6 

7) -10 
8)0,2 

9) ~\ 

10) 0,-10 

11) 

12) -f 

13) -2 
14)0,-i 
15) -8,4 
16)0,1 



17) 
18) 
19) 



7x 



."id 



s - 


Reduce R 


20 


. 7 
' 8fc 


21 


4 

X 


22 


9x 
' ~2~ 


23 


I 3m -4 
' 10 


24 


. 10 


' 9n 2 (9n + 4) 


25 


. 10 

' 2p + l 


26 


1 
' 9 


27 


1 

a; + 7 


28 


. 7m + 3 
' 9 


29 


I 8x 
' 7(x + l) 


30 


, 7r + 8 
' 8r 


31 


, n + 6 
n — 5 


32 


, 6 + 6 
' 6 + 7 


33 


. 9 
' v-W 


34 


. 3(x-3) 

' 5a; + 4 


35 


. 2x-7 
' 5x-7 


36 


fc-8 
' jfe + 4 



37 
38 
39 
40 

41 
42 
43 

44 
45 
46 
47 
48 
49 
50 



3a - 



5a + 2 

9 

P + 2 

2n- 1 
9 

3a; — 5 

5(a; + 2) 

2(m + 2) 
5m — 3 

9r 
5(r + l) 

2(x-4) 

3a; -4 

56-8 



56 + 2 

7n-4 
4 

5(^ + 1) 
3u + l 

6(n + l) 

7a; -6 
(3aT+4)(a7+l) 

7a + 9 

2(3a-2) 

2(2fc + l) 

9(fc-l) 



7.2 

1) Ax 2 

l-i 



2) 



Answers - Multipy and Divide 

3) 
4) 



63 
lOn 

63 

10m 



466 



5) 



3.x 2 



6) 


op 
~2 


7) 


5m 


8) 


7 
To 


9) 


r — 6 
r + 10 


10 


) x + 4 



11 



b 


-5 


X 


-10 




7 




1 


V 


- 10 


X 


+ 1 


a 


+ 10 



a — 6 



17) 5 



18) 



p- 10 

p -4 



19 
20 

21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 



3 

5 

x + 10 



z + 4 


4(m-5) 


5m 2 


7 


z + 3 


4 


n-9 


n + 7 


6 + 2 


86 


u-9 


5 


1 


n — 6 


s+1 


a; — 3 


1 


a + 7 


7 


8(fc + 3) 


x — 4 


z + 3 


9(z + 6) 



10 



33 

34 

35 
36 

37 

38 
39 
40 
41 
42 
43 
44 



9m 2 (m + 10) 

10 



9(n + 6) 


p + 3 


6(p + 8) 


x-8 


x + 7 


56 



6 + 5 

n + 3 
r-8 

18 
~o~ 

3 

2 

1 

a + 26 

1 

x + 2 

3(x-2) 

4(z + 2) 



7.3 

1) 18 

2) a 2 

3) ay 

4) 202?/ 

5) 6a 2 c 3 

6) 12 

7) 2x - 8 

8) x 2 - 2a; - 3 

9) 2 2 -x-12 

10) x 2 -llx + 30 



Answers - Least Common Denominators 



467 



11) 12a 4 6 5 

12) 25x 3 y 5 z 

13) x (x — 3) 

14) 4(x-2) 

15) (s + 2)(x-4) 

16) x(x-7)(x + l) 

17) (x + 5)(x-5) 

18) (s-3) 2 (x + 3) 

19) (a; + l)(:r + 2)(a; + 3) 

20) (a;-2)(x-5)(a: + 3) 

ia 4 26 



21 

22 
23 
24 
25 
26 
27 
28 
29 
30 



10a 3 fc 2 ' 10a 3 6 2 



3x 2 + 6x 


2x-8 


(x-4)(x + 2)' 


(x-4)(x + 2) 


x 2 + 4x + 4 


x 2 - 6x + 9 


(x-3)(x + 2)' 


(x-3)(x + 2) 


5 2x- 


12 -3x 


x(x — 6) ' x(x - 


- 6) ' x(x - 6) 


x 2 — 4x 


3x 2 + 12x 


(x-4) 2 (x + 4) 


' (x-4) 2 (x + 4) 


5x + l 


4x + 8 


(x-5)(x + 2)' 


(x-5)(x + 2) 


x 2 + 7x + 6 


2x 2 - 9x - 18 



(x-6)(x + 6) 2 ' (x-6)(x + 6) 2 

3x 2 + 4x + 1 2x 2 - 8x 



(x-4)(x + 3)(x + l)' (x-4)(x + 3)(x + l) 

4x x 2 + 4x + 4 



(x-3)(x + 2) ' (x-3)(x + 2) 

3x 2 + 15x x 2 -4x + 4 5x-20 



(x-4)(x-2)(x + 5)' (x-4)(x-2)(x + 5)' (x - 4)(x - 2)(x + 5) 



7.4 

Answers - Add and Subtract 
I)- 5 - 

I a + 3 

2) x-A 

3) t + 7 

468 



4) 


a + 4 
a + 6 


5) 


x + 6 

x — 5 


6) 


3x + 4 

~2 



7) 
8) 

9) 
10) 

11) 



24r 

7x + 3y 

X 2y2 

15t+16 
18i 3 

5x + 9 
24 

a + 8 



12) 
13) 
14) 
15) 
16) 
17) 
18) 
19) .,, 



5a 2 + 7a - 3 



9a 2 

- 7x - 13 

4x 

- c 2 + cd - d 2 

Ai 2 

3j/ 2 -3xi/-6x 2 



2x 2 y 2 



1:/' 



x 2 -l 

- z 2 + 5z 

2 2 -l 

lis + 15 

4x(x + 5) 

14 -3a; 



x 2 — 4 

20) i^ 



a; 2 -25 



21 

22) 

23) 

24) 

25) 

26) 

27) 

28) 

29) 

30) 

31 

32) 

33) 



44-5 


4(t-3) 


2x + 10 


(x + 3) 2 


6-20x 


15x(x + l) 


9a 


4(a-5) 


t 2 + 2fy-y 2 


j/ 2 -t 2 


2x 2 - lOx + 25 


x(x — 5) 


x — 3 


(x + 3)(x+l) 


2x + 3 


(x-l)(x + 4) 


x-8 


(x + 8)(x + 6) 


2x-5 


(x-3)(x-2) 


5x+12 


x 2 + 5x + 6 


4x + l 


(x + l)(x-2) 


2x + 4 



34) 


2x + 7 


x 2 + 5x + 6 


35) 


2x-8 


x 2 - 5x — 14 


36) 


- 3x 2 + 7x + 4 


3(x + 2)(2-x) 


37) 


a-2 
a 2 -9 


38) 


2 

y 2 -y 


39) 


2-3 
22-1 


40) 


2 

r + s 


41) 


5(x- 1) 


(x + l)(x + 3) 


42) 


5x + 5 


x 2 + 2x - 15 


43) 


- (x - 29) 


(x-3)(x + 5) 


4 4\ 


5x-10 



x 2 + 4x + 3 



x 2 + 5x + 4 



7.5 

1) 
2) 
3) 
4) 

5) 

6) 

nl 

9) - 
10) 

ID 



x-l 




1-2/ 




J/ 




— a 




a + 2 




5 — a 




a 




a-1 




a + 1 




6 3 + 2b - b - 


-2 



86 



1 

2 

x 2 — x — 1 

X 2 + X + 1 



Answers - Complex Fractions 

2a 2 - 3a + 3 



12 
13 

14 
15 
16 
17 
18 
19 
20 
21 
22 



- 4a 2 - 2a 
x 
3" 

3x + 2 

46(a - 6) 

a 
x + 2 



x-l 
x — 5 



x + 9 

(x-3)(x + 5) 



4x 2 - 


-5x + 4 


1 




3x + 8 




1 




x + 4 




x-2 




x + 2 




x-7 





23) 


X — o 

x + 4 


24) 


-2a -2 
3a -4 


25) 


b-2 
26 + 3 


26) 


x + y 
x-y 


27) 


a — 3b 
a + 36 


28) 


2x 

x 2 + l 


29) 


2 

y 


30) 


x 2 -l 


31) 


y-x 

xy 


?o\ 


x 2 — xj/ + y 2 



x + 5 



y-x 



469 



33) 

34) 



x 2 + y 2 
xy 

2x -1 

2x + l 



35) 
36) 



(l-3x) 2 



x 2 (x + 3)(x-3) 
x + y 



xy 



7.6 



1\ 40 

l) T = a 



Answers - Proportions 



14 

T 

12 

T 



2) n 

3) jfe 

4) £ = 16 
5)x = | 

6) n = 34 

7) 771 = 21 

o\ 79 

8)x = T 

9) p = 49 

10) n = 25 

11) 6 

12) r 5 

13) x ' 

14) n 



40 

~3~ 



36 



2 
32 



15 

16 

17 

18 

19 

20 
21 

22 
23 
24 
25 
26 
27 
28 



a = 


6 
7 


v = 


16 

T 


v = 


69 


n = 


61 
T 


x = 


38 


k = 


73 

IT 


x = 


-8,5 


x = 


-7,5 


m = 


= -7,8 


x = 


-3,9 


P = 


-7,-2 


n = 


-6,9 


n = 


-1 


n = 


-4,-1 



29) 


x = — 7, 1 


30) 


x = — 1,3 


31) 


$9.31 


32) 


16 


33) 


2.5 in 


34) 


12.1 ft 


35) 


39.4 ft 


36) 


3.1 in 


37) 


T: 38, V: 57 


38) 


J: 4 hr, S: 14 hr 


39) 


$8 


40) 


C: 36 min, 




K: 51 min 



7.7 



1) - 


1 2 

2 ' 3 


2) - 


3,1 


3) 3 




4) - 


1,4 


5) 2 





7) - 



Answers - Solving Rational Equations 

8) - 1 - 15) -8 

9) -5 16) 2 

10 )-i i7) -is 

11) -5,0 

12) 5,10 



13) f, 5 

14) 2, 13 



18) -|,1 

19)| 

20) 10 



470 



21) 0, 5 



22) 



2, 



23) 4,7 

24) -1 

25)| 



26)^ 

28) 1 

29) -I 

30) -1 



31) 



13 



32) 1 

33) -10 

34)^ 



Answers - Dimensional Analysis 



1) 12320 yd 

2) 0.0073125 T 

3) 0.0112 g 

4) 135,000 cm 

5) 6.1 mi 

6) 0.5 yd 2 

7) 0.435 km 2 

8) 86,067,200 ft 2 

9) 6,500,000 m 3 

10) 239.58 cm 3 

11) 0.0072 yd 3 

12) 5.13 ft/sec 

13) 6.31 mph 

14) 104.32 mi/hr 

15) 111 m/s 



16 
17 
18 
19 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 



2,623,269,600 km/yr 
11.6 lb/in 2 
63,219.51 km/hr 2 
32.5 mph; 447 yd/oz 
6.608 mi/hr 

17280 pages/day; 103.4 reams/month 
2,365,200,000 beats/lifetime 
1.28 g/L 
$3040 

56 mph; 25 m/s 
148.15 yd 3 ; 113 m 3 
3630 ft 2 , 522,720 in 2 
350,000 pages 
15,603,840,000 ft 3 /week 
621,200 mg; 1.42 lb 



Answers - Chapter 8 



8.1 



1) 7\/5 

2) 5\/5 



Answers - Square Roots 

3) 6 

4) 14 



471 



5) 2y/Z 

6) 6\/2 

7) 6\/3 

8) 20\/2 

9) 48\/2 

10) 56\/2 

11) -112\/2 

12) -21\/7 

13) 8\/3n 

14) 7\/76 

15) Uv 

16) lOn-y^ 

17) 6xy/7 



1) 5^5 


2) 5\/3 


3) 5\/6 


4) 5'V2 


5) 5'V7 


6) 2\/3 


7) -8V6 


8) -16V3 


9) 12V7 


10) 6V3 


11) -2V7 


12) 15a/3 


13) 3V8^ 


14) 2V4n 3 



18 
19 
20 
21 

22 
23 
24 
25 
26 
27 

28 
29 
30 



Answers - Higher Roots 



10aV2a 
-10k 2 

— 20p 2 V7 

— 56x 2 
-16\/2n 

— 30y/m 
32p\fl 
3xy\/5 
66 2 a\/2a 
Axy^/xy 

16aW2 

8x 2 y 2 \fb 
16m 2 n\/2n 



15) 


2\flrf 


16) 


-2^3^ 


17) 


2pV7 


18) 


2xVi 


19) 


— <o\flr 


20) 


-166V36 


21) 


Av 2 V6v 


22) 


20a 2 V2 


23) 


-28n 2 V5 


24) 


-8n 2 


25) 


— 3xyv5x 2 


26) 


Auvvu 


27) 


— 2xy\/4xy 


28) 


lOabV^ti 2 


29) 


Axy 2 V4x 



31 
32 
33 
34 
35 
36 
37 
38 
39 
40 
41 
42 



30 
31 
32 

33 

34 

35 
36 
37 

38 
39 
40 
41 



24t/V5a; 
56\/2mn 



35xyv5?/ 
12xy 

— 12u\/ouv 

— 30y 2 x^/2x 

— 48xVy\/5 
30o 2 c\/2fe 

■■\ 

Ayzy2xz 
12p\/6mn 
32p 2 m^2q 



<,j 2 V5hk 



3xy 2 %/7 
-2lxy 2 %/3y 



-8y 2 y7x 2 y 2 


10v 2 V3u 2 v 2 


-40y6xy 


- l2%/3ab 2 


9y'V5x 


— 18m 2 np 2 \j2m 2 p 


4 / q 

— Vlmpqyhp 


18xy\/8xy 3 z 2 


-18ab 2 i/5ac 

„_ „ A 1 — —R 



Uhfk 2 V8h 2 



472 



42) -18xz\/4x 3 yz 3 




8.3 






Answers - Adding Radicals 


1) 6\/5 


21) -4\/6+4\/5 


2) -3\/6-5\/3 


22) -\/5-3\/6 


3) -3\/2 + 6\/5 


23) 8\/6-9\/3+4\/2 


4) -5\/6-\/3 


24) -\/6-10\/3 


5) -5\/6 


25) 2^2 


6) -3\/3 


26) 6^5-3^3 


7) 3\/6 + 5\/5 


27) -V3 


8) -\/5 + \/3 


28) 10Vi 


9) -8\/2 

10) -6\/6+9\/3 

11) -3\/6 + \/3 

12) -2\/5-6\/6 


29) V2-3V3 

30) 5V6 + 2V4 
31)6V3-3VI 


13) -2\/2 


32) -6V3 + 21/6 


14) 8\/5-\/3 


33)2V2+V3 + 6V4 


15) 5\/2 


34) -2V3-9V5-3V2 


16) -9\/3 


35) \/6-6\/2 


17) -3\/6-\/3 


36) V3 - 6V6 + 3\/5 


18) 3\/2 + 3\/6 


37) 4\/5 - 4^6 


19) -12\/2 + 2\/5 


38) -llV2-2\/5 


20) -3\/2 


39) _ 46/4 - 6\/5 - 4V2 


8.4 






Answers - Multiply and Divide Radicals 


1) -48\/5 


5) 2x 2 \fx 


2) -25\/6 


6) 6a 2 Voa 


3) 6mv5 


7) 2\/3 + 2\/6 


4) -25r 2 \/2r 


8) 5\/2 + 2\/5 



473 



9) - 45\/5 - Hh/15 



10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
20 

21 
22 

23 

24 



45\/5 + 10v / 15 

25Wl0 + 10<\/5 

5\/3-9\/5v 

-2-4\/2 

16-9V3 

15-ll\/5 

30 + 8\/3 + 5\/l5 + 4\/5 

6a + a\/T0 + 6a\/6 + 2a\/l~5 

- 4p\/T0 + 50^/p 
63 + 32\/3 

- 10^/m + 25\/2 + \/2m - 5 
VI 

25 

\/l5 

4 



1 

20 



ZUJ 


3 


26) 


%/io 

15 


27) 


4^3 
9 


28) 


4^5 
5 


29) 


5v / 3Sj/ 


122/2 


30) 


4^3~r 
16xy 2 


31) 


3 


32) 


2^5ri 
5 


33) 


5 




^15 



35) 
36) 



10 



l vt 



Answers 



2) 
3) 
4) 

5) 
6) 

7) 
8) 



4 + 2^3 
3 

-4 + ^3 
12 

2 + V3 



5 



2^13-5^65 
52 

^85+4^17 
68 

^6-9 
3 

30-2^3 
~1~8 



U, J 


2 


38) 


V4n 2 
mn 


Rationalize Denominators 


qx 15V^-5V2 
^ 43 


10) 


- 5^3 + 20^5 

77 


11) 


10-2^2 
23 


12) 


2^3 + ^ 
2 


13) 


- 12-9^3 
11 


14) 


-2\/2-4 


15) 


3-Vb 


161 


V5-V3 



17) l + \/2 



474 



■|M 16^3+4^5 „qN a 2 - 2aVb + b 

'43 ' a 2 — b 

19)\/2-l 30)Va-\/6 



20)3 + 2\/3 31)3\/2 + 2\/3 



21) \/2 32) a ^ + b ^ 



33) 



22) V2 

23) V^ 

24) 3-2\/2 



34) 



a — b 




ayo + 6\/a 




ab 




24-4^6 + 9\/2- 


-3^3 



35) 
26) i 36 ) 



15 
1 + ^5 



25) V^ 4 

2^5-572-10 + 5^10 
30 

-5a/2 + 10-a/3+\/6 



3 



27) 4 - 2\/3 + 2\/6 - 3\/2 37) 

O0 x 2^-2^15 + ^3+3 s 8 + 3^ 

28 ) ~2 38 ) -~W~ 



5 



Answers - Rational Exponents 



1) (^/mf 
n \ i 



' (VW) 3 
3) (y/Tx) 

4) 



i 



10) 2 

11) 8 



iu; 


i i 

a 2 b 2 


16) 


1 


17) 


1 
3^ 


18) 


25 
j,12 

5 




.X 6 


19) 


11 


20) 


1 


21) 


1 


22) 


u 2 
7 




• U 2 


23) 


7 3 
b 4 aA 

3 


24) 


17 

2y^ 

7 




X 4 


9M 


1 

3yi2 





3 


26) 


a 2 

r 




264 




35 


27) 


m 8 

7~~ 




n 6 


28) 


1 

5 3 




y 4 x 2 



(VeE) 4 
5) (6x)"^ 

_ 7 

7) n ¥ 



8) (5a)* ^^ 30) 

9) 4 



y 1 



31) x^ 



4 

12 ) KXXJ ....if 32) 4 

4 5_ 
13) X3y2 

14) 4 ok n 3„*r 33) - 



475 



34) x 4 y 



8.7 



i) vw 

2) yW 

3) V«a: 2 2/V 



4) 
5) 



2x 

V35xy 
3y 



6) v»v? 



7) v^v 



5/g^l.,^ 



a y 



9) 4 aAV^ 



10 
11 
12 
13 

14 
15 
16 
17 
18 
19 
20 
21 
22 
23 



5y 
V2xV 2 

vw 

V5400 
1 v / 300125 

Vx 3 (x - 2) 2 
V3x(z/ + 4) 2 



10/ 9 7 

van/ 



i^TTTm 



x"2r 



20, 



20/ 
V 



a 



',17 



-°/' ]S I( 17 c 14 



- Radicals of Mixed Index 


24 


) ^x 22 y u z 27 


25 


) ai/a 


26 


) Xy/x 


27 


) bvo 


28 


i2rT? 
av a 


29 


) xyyxy* 


30 


. 10/— T7 

a v ab 


31 


3a 2 bvab 


32 


) 2xy 2 %j2x b y 


33 


) x ^59049a; y ll z 10 


34 


) a 2 b 2 c 2 va 2 b c 2 


35 


) 9a 2 (6+l)V243a 5 (6 + l) 5 


36 


) Ax(y + zf \/2x(y + z) 


37 


i2r^ 
) Va 5 


38 




39 


) 1 v / ^V 


40 


' b 


41 


) 7a 6 V 


42 


) yz 1 \/xy 8 z 3 


43 


)'^(Sx-lf 


44 


) 1 V / (2 + 5x) 5 


45 


(V^ + l) 4 


46 


) V(5-3^) 



476 



Answers - Complex Numbers 



1) 


11 — 4i 


2) 


-4z 


3) 


-3 + 9i 


4) 


— 1 — 6i 


5) 


— 3 — 13« 


6) 


5-12i 


7) 


-4-lli 


8) 


— 3 — 6« 


9) 


-8-2i 


10) 


13 -8i 


11) 


48 


12) 


24 


13) 


40 


14) 


32 


15) 


-49 


16) 


28-21z 


17) 


ll + 60z 


18) 


— 32 — 128i 


19) 


80-10z 


20) 


36-36? 


21) 


27 + 38z 



22 


) - 28 + 76i 


23 


) 44 + 8i 


24 


) 16- 18i 


25 


) -3+lli 


26 


) -l + 13i 


27 


) 9i + 5 


28 


( -3i-2 

' 3 


29 


. 10i-9 

' 6 


30 


( 4i + 2 
' 3 


31 


. 3i-6 

' 4 


32 


. 5i + 9 
' 9 


33 


) 10i + l 


34 


) -2i 


35 


l -40i + 4 
' 101 


36 


( 9i-45 
' 26 


37 


. 56 + 48i 
' 85 


38 


( 4-6i 
' 13 


39 


, 70 + 49J 
' 149 


40 


. -36 + 27i 



41 
42 
43 
44 
45 
46 

47 

48 
49 

50 
51 

52 
53 
54 
55 
56 
57 
58 



30i - 5 



37 


48i - 56 


85 


9/ 


3z\/5 


-2y/5 


-2\/6 


l + iV3 


2 


2 + iV2 


2 


2-i 


3 + 2iV2 



Answers - Chapter 9 



9.1 

1)3 

2) 3 

3) 1, 5 

4) no solution 

5) ±2 



Answers - Solving with Radicals 

6) 3 11) 6 

8) no solution 

9) 5 

10) 7 14) 21 



12) 46 

13) 5 



477 



15) 



16) 



9.2 



1) 


±5\/3 


2) 


-2 


3) 


±2\/2 


4) 


3 


5) 


±2\/6 


6) 


-3,11 


7) 


-5 


8) 


1 3 

5' ~~ 5 



Answers - Solving with Exponents 



10) 



l±3\/2 



9) 



11) 65, -63 

12) 5 

13) -7 
14) 

15) 

16) 

17) 



_ 11 5 

~ ~2~' 2 
11 _ 5 
~2~' _ 2 

_ 191 
64~ 

_ 3 _ 5 
~~ 8' ~~ 8 



18) 


9 

8 


19) 


5 

4 


20) 


No Solution 


21) 


-£-10 


22) 


3 


23) 


17 
2~ 


24) 


No Solutoin 



9.3 







Answ( 


:rs - Complete the Square 


1) 225 


(x-15) 2 




17) - 5 + a/86, - 5 - \/86 


2) 144 


(a-12) 2 




18) 8 + 2\/29,8-2\/29 


3) 324 


(ra-18) 5 




19) 9,7 


4) 289 


(x-17) 2 




20) 9,-1 


5)?;(x-^) 2 




21) - 1 + iy/21, - 1 - iy/21 


6 )^( r "i) 2 




22) 1,-3 


7)*(v-*) a 




21) 1—1 

Z6 > 2' 2 


8)T;b-¥) 2 




24) 3,-1 


9) 11,5 




25) -5 + 2z,-5-2z 


10) 4 + 2\/7,4- 


2\/7 


26) 7 + \/85 , 7 - \/85 


11) 4 + i\/29,4- 


i\/29 


27) 7,3 


12) -l + n/42, - 


-l-i\/42 


28) 4,-14 


1q n 1 -2 + i^ -2- 


-i\/38 

2 


29) l + iy/2, l-iy/2 


-..} 3 + 2i%/33 3-2« 
' 3 ' 3 


n/33 


QfA 5 + i^l05 5-^^/l05 
6() ) 5 ' 5 


15) 5 + 


i-v/215 5-i-v 


/215 


01 n 4 + iVHO 4- iVHO 


5 ' 5 




61 > 2 ' 2 


16) " 4 


+ 3\/2 -4- 
4 ' 4 


3^2 


32) 1,-3 



478 



33) 4 + iV39,4-zV39 

34) -1.-7 

35) 7,1 

36) 2,-6 

-6 + ^^/258 - 6 - iV258 



6 ' 6 

-6 + i^TTT -6-n/ITT 



37) 

38) 

39 ) 5 ' 5 

40) 2,-4 

-5 + i^ -5-i\/87 



3 ' 3 

5 + i^l30 5-«^T30 



41) 
42) 
43) 
44) 



2 ' 2 

-7 + a/181 -7-VT81 



2 ' 2 

3 + i^27l 3-«^27l 



7 ' 7 

l + 2i^6 -l-2i-v/6 



45 
46 
47 
48 
49 
50 
51 
52 

53 

54 
55 

56 



7 + iV139 7-iV139 



2 ' 2 

5 + 3«^7 5 — 3i\/7 



12 



,-4 



1 + JV5TT l-iV511 



4 ' 4 

9 + ^21 9-\/21 

2 ' 2 

l + n/l63 1-J163 



2 ' 2 

-5 + 4^415 -5-i\/415 



11 + JV95 11-JV95 



5 + JV19T 5-JV191 



9.4 



' "2" ' 2~~ 








3) 2 + V5 , 2 - \/5 




4 x %/6 v/6 
' 6 ' 6 




5 ) -2"'-^- 




~x -l + iV29 -1-iv 
6 ) 5 > 5 


/29 


7 )W 




Q x l + \/3T l-x/31 
8 ) 2 ' 2 




9) 3,-3 




10) iV2,-iV2 




11)3,1 




12) -l + i,-l- 


i 


-1 q^ — 3 + i-\/55 —3 — 
irf J 4 ' 4 


^55 


1 4"\ -3 + i^l59 -3- 
i4 / 12 ' 


^^/l59 
12 



Answers - Quadratic Formula 

•3 + -/T4T -3--/141 



15) 

16) y/S,-y/3 



6 



-3 + V401 -3-V401 



14 ' 14 

■5 + ^137 -5--\/l37 



17) 

18 ) 8 > 8 

19) 2,-5 

20) 5,-9 

r)-,\ — l + iv/3 - 1 - i\/3 



22)3,-1 

23)^,-7 



24) 
25) 
26) 
27) l(i 



3 + i^3 - 3 - i^3 



3 ' 3 

7 + 3^21 7-3\/2l 
10 ' 10 

-5 + ^337 -5-^337 



12 ' 12 

3 + i\/247 -S-iv 7 ^ 



16 



479 



oo\ 3 + V33 

28) 6 


3- V33 
6 


29) -1,- 


3 
2 


30) 2\/2,- 


2a/2 


31) 4,-4 




32) 2,-4 




33) 4,-9 




om 2 + 3iV5 


2-3«^5 



35) 6,-| 

o fi x 5 + i^l43 5-i\/l43 
37) 



38) 

39) 

40) 



14 


5 


14 




-3 + v 


/ 345 - 


3- v 


/ 345 


14 


; 


14 




>/6 

2 ' 


>/6 

2 






726 


a/26 

2~~ 






-1 + v 


/ 141 - 


1- V 


/ 14T 



10 



10 



9.5 

Answers - Build Quadratics from Roots 

NOTE: There are multiple answers for each problem. Try checking your answers 
because your answer may also be correct. 

1) 2 2 - 72 + 10=0 

2) x 2 - 9x + 18 = 

3) x 2 -22x + 40 = 

4) £ 2 - 14x + 13 = 

5) x 2 - 8x + 16 = 

6) x 2 -9x = 

7) x 2 = 

8) x 2 + 7x + 10 = 

9) x 2 - 7x - 44 = 

10) 2 2 - 22 -3 = 

11) I62 2 - 162 + 3 = 

12) 562 2 - 752 + 25 = 

13) 62 2 -52 + l = 

14) 62 2 - 72 + 2 = 



15 


) 72 2 -312 + 12 


= 


29) 


x 2 +13 = 


16 


) 92 2 - 202 + 4 = 


= 


30) 


x 2 + 50 = 


17 
18 
19 


) 182 2 -92-5 = 
) 62 2 — 72 — 5 = 
) 92 2 + 532-6 = 


= 


= 


31) 
32) 


2 2 - 42 - 2 = 
2 2 + 62 + 7 = 


20 


) 52 2 + 22 = 




33) 


2 2 - 22 + 10 = 


21 


) 2 2 - 25 = 




34) 


2 2 + 42 + 20 = 


22 


) 2 2 - 1 = 




35) 


2 2 - 122 + 39 = 


23 


) 252 2 -l = 




36) 


2 2 + 182 + 86 = 


24 
25 


) 2 2 - 7 = 

) 2 2 -ll = 




37) 


42 2 + 42 - 5 = 


26 


) 2 2 -12 = 




38) 


92 2 - 122 + 29 = 


27 


) 162 2 -3 = 




39) 


642 2 - 962 + 38 = 


28 


) 2 2 + 121 = 




40) 


42 2 + 82 + 19 = 



9.6 

1) ±1,±2 

2) ±2,±VE 

3) ±i,±2y/2 

4) ±5, ±2 



Answers - Quadratic in Form 

5) ±1,±7 

6) ±3,±1 

7) ±3, ±4 

8) ±6, ±2 



480 



9) 


±2, ±4 


10 


)2,3, l±n/3, - 3± 2 3n/5 


11 


) 2,3,l±n/3, - 3 ± i73 


12 


) ±\/6,±2i 


13 


. ±2«^ ±V6 
' 3 ' 2 


14 


1 1 
' 4' _ 3 


15 


) -125,343 


16 


. 5 1 

' ~~ 4 ' 5 


17 


. -, 1 l±i^3 -l±iV3 
) J-) 2 ' 4 ' 2 


18 


) ±2,±\/3 


19 


) ±i,±\/3 


20 


) ±i\/5,±i\/2 


21 


) ±A±^ 


22 


, . ±6 
) ±1,^- 


23 


) ±l,±2\/2 


24 


)2,V2, lii^,"^^ 


25 


. -, 1 -l±i^3 -l±i\/3 
' X ' 2' 4 ' 2 


26 


. 1 -, -l±i^3 l±i^3 
' 2' ' 4 '2 



27^ 


±l,±i,±2, 


±2i 


28; 


6, 




29; 


-(6+3), 7- 


■6 


30; 


-4 




31; 


-4,6 




32; 


8,-1 




33; 


-2,10 




34; 


2, -6 




35; 


-1,11 




36; 


§.° 




37; 


4.-1 




38; 


±\/6,±\/2 




39; 


±1 -1 5 - 

' 3 ' 3 




40; 


0,±l,-2 




41; 


511 1339 
3 ' 24 




42; 


-3, ±2,1 




43; 


±1,-3 




44; 


— 3 — 1 - - 

-3, J-, 2 , 


1 
2 


45; 


±1 -1 * 
' 2' 2 




46; 


i)Z ' 3' 3 





9.7 

1) 6 m x 10 m 

2) 5 

3) 40 yd x 60 yd 

4) 10 ft x 18 ft 

5) 6 x 10 

6) 20 ft x 35 ft 

7) 6" x 6" 

8) 6 yd x 7 yd 

9) 4 ft x 12 ft 



Answers - Rectangles 

10) 1.54 in 

11) 3 in 

12) 10 ft 

13) 1.5 yd 

14) 6 m x 8 m 

15) 7 x 9 

16) 1 in 

17) 10 rods 

18) 2 in 



19) 15 ft 

20) 20 ft 

21) 1.25 in 

22) 23.16 ft 

23) 17.5 ft 

24) 25 ft 

25) 3 ft 

26) 1.145 in 



481 



9.8 

1) 4 and 6 

2) 6 hours 

3) 2 and 3 

4) 2.4 

5) C = 4, J = 12 

6) 1.28 days 

7) l| days 

8) 12 min 

9) 8 days 

10) 15 days 



Answers - Teamwork 

11) 2 days 

12) 4^ days 

13) 9 hours 

14) 12 hours 

15) 16 hours 

16) 7- min 

17) 15 hours 

18) 18 min 

19) 5- min 

20) 3.6 hours 



21) 24 min 

22) 180 min or 3 hrs 

23) Su = 6, Sa = 12 

24) 3 hrs and 12 hrs 

25) P = 7, S = 17| 

26) 15 and 22.5 min 

27) A = 21, B = 15 

28) 12 and 36 min 



9.9 









Answers - Simultaneous Product 


1)(2,36),(- 


18,- 


-4) 


7) (45, 2), (-10, -9) 


2) (-9,-20), (- 


40,- 


|) 8) (16,3), (-6,-8) 


3) (10,15),( 


-90, 


--) 

3' 


9) (1,12), (-3, -4) 


4)(8,15),(- 


10,- 


-12) 


10) (20, 3), (5, 12) 


5) (5,9), (18, 


2.5) 




11) (45,1), (-|,-27) 


6) (13,5),(- 


20,- 


--) 

4 / 


12) (8, 10), (-10, -8) 


9.10 






Answers - Revenue and Distance 


1) 12 






9) 60 mph, 80 mph 17) r = 5 


2) $4 

3) 24 






10 > 60 ' 80 18) 36 mph 

11) 6 km/hr 


4) 55 






12) 200 km/hr 19 ) 45 m P h 


5) 20 






13 ) 56 > 76 20) 40 mph, 60 mph 


6) 30 

7) 25 @ $18 






14) 3.033 km/hr 

ir , in . nA . 21) 20 mph 

15) 12 mph, 24 mph ; F 


8) 12 @ $6 






16) 30 mph, 40 mph 22) 4 mph 



9.11 



482 



Answers - Graphs of Quadratics 




2) 



(-i,o\ 

(0, -3) 



K3,0) 

j 

(1, -4) 



3) 



(0,10) 

(1,01 (5.0) 



(3,-8) 




(0,-18) 



6) 



(M) 

(-10,0) 



(3,8) 
4X5.0) 



7) 



(3,0)/ \( 5 ,0) 



(0,-45) 



V44,3j 



8) 



-M£ 



9) 



-9,0) 



(0,5) 

(-Ml 



(2,3) 



43,0) 



(2,9) 



45,0) 



10) 



(1,0) , C2,l) 



(0,-3) 



T3,0) 



11) 



4141} 



(0,-5) 



(3,4) 



P.O) 



12) 



43411 



(0,-30) 



44,2) 
T45,0) 



13) 



44,8) 
(2,0)/ \(6,0) 



(0,-24) 
/ 

14) 

(- 3,o\ 

(-1,-8^ 

15) 



/ (i,Q) 

(0,-6) 



(-3,0)\ (-1,0), 



/ (0,9) 



(-2,-34 



16) 



(0,45) 
(-3,0) 



17) 



(0,75) 



(- 3,0) 

(-2,-5) 



\(3,0)/ (5 ,0) 
(4,-5) 

(0,15) 

T4oT 



483 



19) 



(-7,0) 




(0,-175) 



20) 



W 



(-15,0) 



N (2,5) 
(3,0) 



Answers - Chapter 10 



10.1 



Answers - Function Notation 



1) a. yes b. yes c. no 
d. no e. yes f. no 
g. yes h. no 



2) 


all real numbers 


3) 


S- 5 


4) 


t^O 


5) 


all real numbers 


6) 


all real numbers 


7) 


x ^ 16 


8) 


x^-1,4 


9) 


x ^ 4, x ^ 5 


10) x^±5 


11 


) "4 


12 


) -- 

' 25 



13) 2 



14) 


85 


15) 


-7 


16) 


7 


17) 


17 


18) 


-6 


19) 


13 


20) 


5 


21) 


11 


22) 


-21 


23) 


1 


24) 


-4 


25) 


-21 


26) 


2 


27) 


-60 


28) 


-32 



29) 
30) 


2 

31 

32 


31) 


- 64a; 3 + 2 


32) 


4n + 10 


33) 


-l + 3x 


34) 


12 + a 

-3-2 ^ 


35) 


2 -3n 2 -l 


36) 


! + > 2 


37) 


3x + l 


38) 


t 4 + t 2 


39) 


5-3-x 


40) 


-2 + n 

5 2 +1 



10.2 

1) 82 

2) 20 

3) 46 

4) 2 

5) 5 



Answers - Operations on Functions 

6) -30 

7) -3 

8) 140 
9)1 

10) -43 



484 



n; 


100 


12; 


-74 


13; 


1 

5 


14; 


27 


!s; 


9 

26 


16; 


n 2 — 2n 


!7; 


- x 3 - Ax - 2 


is; 


-2 3 + 22 2 -3 


19; 


-2 2 -82 + 2 


20; 


2i 2 - St 


21; 


4x 3 + 25x 2 + 252 


22; 


-2t 3 -15t 2 -25t 


23; 


x 2 — 4x + 5 


24; 


32 2 + 4a; - 9 


25; 


n 2 + 5 
3n + 5 


26; 


-22 + 9 


27; 


-2a + 5 
3a + 5 


28; 


t 3 + 3t 2 - 3« + 5 


29; 


n 3 + 8n + 5 


30; 


4x + 2 
a: 2 + 2a: 


31; 


n 6 - 9n 4 + 20n 2 


32; 


18n 2 -15n-25 


33; 


2 + 3 


34; 


2 
~~ 3 


35; 


t 4 + 8t 2 + 2 



36 

37 
38 
39 

40 
41 

42 
43 
44 
45 
46 
47 
48 
49 
50 
51 
52 
53 
54 
55 
56 
57 
58 
59 
60 



3n - 


-6 


— n 2 - 


- An 


-Z 3 - 


-2x 



-3X + 4 

x A - 4x 2 - 3 

- n 2 - 2n 



32 + 23n - n 3 

8 

-155 

5 

21 

4 

103 

12 

-50 

112 

176 

147 

I62 2 + 12x-4 

-8a + 14 

-8a + 2 

t 

4x 3 

-2n 2 -12n-16 

-22 + 8 

27t 3 -108t 2 + 141t-60 

-16* -5 

32 3 + 62 2 -4 



10.3 



1) Yes 



Answers - Inverse Functions 

2) No 



485 



3) Yes 

4) Yes 

5) No 

6) Yes 

7) No 

8) Yes 

9) Yes 

10) No 

11) f~\x) 

12) g-\x) 

13) g~\x) 

14) f~\x) 

15) f~\x) 

16) g~ l (x) 

10.4 



1)0 

2) -1 

3) 

4) 

5 ) -! 

6) -f 

7 ) -I 

8) 

9) -! 



:V^^3+2 

(x-2) 3 -l 





4 


-2a; 






X 






3 + 3a; 






X 






2x-2 



x + 2 



3x-9 



10) 

12) 

13) -2 

I 4 )"! 



17) r 1 
is) r 1 

19) (7" 1 

20) r 1 

21) /- 1 

22) g- 1 

23) 4" 1 

24) J" 1 

25) r 1 

26) /i" 1 

27) g~ l 

28) 2" 1 

29) s" 1 



-5^ + 10 

15 + 2x 

/x + 1 

- 4a; + 12 
3 

\fx + Z 
- 2x 5 + 2 





x-1 




— 3a; — 3 




x + 2 




-x-1 




x-1 




-2x 




x-1 




-4a; + 8 




5 


= 


-3a; + 2 




-x + l 



30 
31 
32 

33 

34 
35 

36 

37 

38 
39 
40 



Answers - Exponential Functions 



15) 


1 




16) 




1 


17) 


No solution 


18) 




1 
3 


19) 




1 
4 


20) 




3 
4 


21) 


No solution 


22) 







23) 




3 
2 


24) 


2 
5 




25) 




1 


26) 


1 

4 




27) 




1 
2 


28) 


1 
3" 





29 
30 
31 
32 

33 
34 
35 
36 
37 
38 
39 
40 



f~\x) 



5 + 4a; 



g- l (x)=\/x + l 



■x + 3 



,- 1(x) = (^±il! 

^" 1 (x) = V^ 3 2 + l 

rV) = " 2x+1 

/"V) = 

rV) 2 * +7 

f-\x) 

g~\x) 
g-\x) 



x-1 
-1-x 



x + 3 

Ax 
3~ 

— X 

-3a- + l 





No solution 

1 

3 

1 
3 

2 
3 




3 

8 
-1 

-3 

No solution 



486 



10.5 



Answers - Logarithmic Functions 



1) 9 2 = 81 

2) b~ 16 =a 

3)7" 2 = i 

4) 16 2 = 256 

5) 13 2 =169 

6) 11°= 1 

7) log 8 1 = 



8) logi 7 



i 

289 



9) logis 225 = 2 

10) l0g 144 12 = 1 

ll)log 64 2 = i 

12) log 19 361 = 2 

13)1 

14) 3 



15) 


i 

~~ 3 


16) 





17) 


2 


18) 


-3 


19) 


2 


20) 


l 

2 


21) 


6 


22) 


5 


23) 


5 


24) 


512 


25) 


i 

4 


26) 


1000 


27) 


121 


28) 


256 



29) 6552 

30) § 
31) 
32) 
33) 
34) 
35) 
36) 
37) 
38) 

39) 5 

40) 3 



125 
~3~ 

1 
I 

54 

IT 

2401 



1 
2 

1 

tt 

621 

To" 



283 
243 

2 



10.6 

1) 

a. 740.12; 745.91 

b. 851.11; 859.99 

c. 950.08; 953.44 

d. 1979.22; 1984.69 



Answers - Interest Rate Problems 



e. 1209.52; 1214.87 

f. 1528.02; 1535.27 

g. 2694.70; 2699.72 
h. 3219.23; 3224.99 



i. 7152.17; 7190.52 



2) 1640.70 

3) 2868.41 

4) 2227.41 

5) 1726.16 

6) 1507.08 



7) 2001.60 

8) 2009.66 

9) 2288.98 

10) 6386.12 

11) 13742.19 



12) 28240.43 

13) 12.02; 3.96 

14) 3823.98 

15) 101.68 



487 



10.7 



Answers - Trigonometric Functions 



1) 


0.3256 


2) 


0.9205 


3) 


0.9659 


4) 


0.7660 


5) 


7 

25 


6) 


8 
15 


7) 


7 
16 


8) 


3 

5 


9) 


V2 
2 


10 


>! 


11 


) 16.1 


12 


) 2.8 


13 


) 32 



14) 


8.2 


15) 


26.1 


16) 


16.8 


17) 


2.2 


18) 


9.8 


19) 


17.8 


20) 


10.3 


21) 


3.9 


22) 


10.6 


23) 


10.2 


24) 


8.9 


25) 


9.5 


26) 


24.4 


27) 


4.7 



28) 


14.6 


29) 


1 


30) 


8 


31) 


1.5 


32) 


7.2 


33) 


5.5 


34) 


2 


35) 


41.1 


36) 


3.2 


37) 


18.2 


38) 


3.3 


39) 


17.1 


40) 


22.2 



10. 





Answers - Inverse Trigonometric Functions 


1) 29° 


11) 


36° 


2) 39° 


12) 


61.7° 


3) 41° 


13) 


54° 


4) 52° 


14) 


46.2° 


5) 24° 


15) 


55.2° 


6) 32° 


16) 


42.7° 


7) 15° 


17) 


58° 


8) 18° 


18) 


20.1° 


9) 27° 


19) 


45.2° 


10) 35° 


20) 


73.4° 



488 



21) 51.3° 

22) 45° 

23) 56.4° 

24) 48.2° 

25) 55° 

26) 30.5° 

27) 47° 

28) 15.5° 

29) 30° 

30) 59° 

31) mZB = 28°, b= 15.1, c= 32.2 



32) mZB = 22.8°, mZA = 67.2°, 

c = 16.3 

33) mZB = 22.5°, mZv4 = 67.5°, c = 7.6 

34) mZA = 39°, 6 = 7.2, a = 5.9 

35) mZE = 64.6°, mZA = 25.4°, fr = 6.3 

36) mZi = 69°, b = 2.5, a = 6.5 

37) mZB = 38°, 6 = 9.9, a = 12.6 

38) mZB = 42°, 6 = 9.4, c= 14 

39) mZA = 45°,6 = 8,c=11.3 

40) mZB = 29.1°, mZA = 60.9°, 
a = 12.2 



489